Magnetism, Magnetic Field Force, Right Hand Rule, Ampere's Law, Torque, Solenoid, Physics Problems

01:22:40
https://www.youtube.com/watch?v=csMqfwJRjCs

Résumé

TLDRThe video delves into the principles of magnetism, explaining how magnetic poles interact, such as north-north repelling and north-south attracting. It describes how bar magnets possess magnetic fields that emerge from the north pole and enter the south pole. The video further explains that magnetic fields are generated by moving electric charges, with demonstrations using wires carrying currents. It introduces the right-hand rule for understanding the direction of magnetic fields and covers various formulas for calculating magnetic field strengths and forces on wires and charges, such as B = μ₀I/2πr. Additionally, the video walks through complex examples and equations related to forces acting on conductors in magnetic fields, the behavior of charged particles in magnetic environments, and scenarios like using Ampere's Law and solenoids to examine magnetic effects.

A retenir

  • 🧲 Like poles repel, opposite poles attract.
  • 🔋 Magnetic fields arise from moving charges.
  • 👉 Right-hand rule helps determine magnetic field direction.
  • 📏 Magnetic field decreases with distance.
  • ⚡ Magnetic force is greatest when current and field are at right angles.
  • 🔄 Magnetic forces cause rotating motion in loops.
  • 🖉 Formula B = μ₀I/2πr defines wire's magnetic field.
  • ℹ️ An increase in current increases magnetic field strength.
  • ✖️ Perpendicular forces cause changes in particle motion.
  • 📐 Magnetic dipole moment is NIAB sin θ.

Chronologie

  • 00:00:00 - 00:05:00

    This video begins with an explanation of basic magnetic principles using bar magnets. It describes how like poles (north-north or south-south) repel, while opposite poles (north-south) attract each other. Each bar magnet has its own magnetic field that travels from the north pole to the south pole, and when magnetic fields align from north to south, they add up and cause attraction. Magnetic fields are created by moving electric charges, exemplified by a current-carrying wire which generates a circular magnetic field around it. The right-hand rule is introduced for determining the direction of this magnetic field.

  • 00:05:00 - 00:10:00

    The discussion continues with the mathematical relationship between current, distance, and magnetic field strength around a wire. The formula B = μ₀I/(2πr) is explained, where increasing the current increases the strength of the magnetic field, and increasing the distance from the wire decreases the field strength. The number of magnetic field lines depicted indicates field strength. A practical problem involving the calculation of the magnetic field two centimeters from a wire carrying a current is solved, showing how magnetic fields weaken with distance.

  • 00:10:00 - 00:15:00

    Next, another practical problem is presented, involving calculating the distance from a wire where a specific magnetic field strength will be produced. The solution involves rearranging the formula to solve for distance, emphasizing the relationship between current, field strength, and distance. Additionally, a compass deflection experiment is briefly mentioned to illustrate the presence of a magnetic field around a current-carrying wire. Following this, the video describes how a current-carrying wire inside a magnetic field experiences a force, introducing an equation F = ILBsinθ for magnetic force on a wire and showing dependency on current, field strength, wire length, and angle.

  • 00:15:00 - 00:20:00

    The explanation of magnetic force on a wire continues with examples showing different angles between current and magnetic field direction, highlighting that maximum force occurs when they are perpendicular. Directions of magnetic force are determined using the right-hand rule, and practical problems are solved using given scenarios of current direction, magnetic field orientation, and calculation of force magnitude and direction. The concept that without perpendicular interaction, there's no magnetic force is reinforced through these illustrations.

  • 00:20:00 - 00:25:00

    There's a shift to discussing the magnetic force acting on a moving charge within a magnetic field, transitioning from wires to single point charges. The derived formula F = Bqvsinθ calculates the force on moving charged particles, with specific focus when magnetic force reaches its maximum when velocity and field direction are perpendicular. Examples with protons and electrons demonstrate how directionality of force changes with charge polarity, and calculating the magnitude of such forces illustrates practical application.

  • 00:25:00 - 00:30:00

    The principle of magnetic force causing charged particles to move in circular orbits is introduced, with a derivation of the formula for the radius of such a path (r = mv/Bq), reflecting a balance between centripetal and magnetic forces. A demonstration involving protons moving perpendicular to a magnetic field illustrates the path a charged particle takes within these constraints, and the effects are contrasted with electrons, acknowledging differences due to opposite charge.

  • 00:30:00 - 00:35:00

    Equations are used to calculate the radius of curvature for specific conditions involving charged particles. A practical example leads to using known properties of protons in a magnetic field to find this radius. The energy of such particles is contextualized using electron volts as units, demonstrating the conversion from joules to electron volts reinforcing the understanding of energy in atomic-scale particles. The video transitions into explaining why one electron volt equals 1.6*10^-19 joules in terms of electric potential.

  • 00:35:00 - 00:40:00

    The interaction between two parallel current-carrying wires is explained, showing attraction when currents are parallel and repulsion when they are in opposite directions, deriving from the magnetic fields generated. This section explains how wire-generated magnetic fields affect nearby wires and how to compute the force using derived formulas. A practical example involving two parallel wires is solved to illustrate these principles, including computing forces when currents flow either in parallel or opposite directions.

  • 00:40:00 - 00:45:00

    Using Ampère's law, the video explains deriving the magnetic field around a wire, introducing the relationship between current and magnetic field along a closed loop integral. Essential concepts include envisioning the magnetic field as following a circular path around a wire, leading to deriving the equation B = μ₀I/(2πr). This section emphasizes the significance of current enclosed by the loop in shaping magnetic field distribution, rounding off with the typical scenarios exhibited in practical applications.

  • 00:45:00 - 00:50:00

    Ampère’s law is applied further to derive expressions for magnetic fields in solenoids, explaining how the cumulative effect of multiple loops intensifies the field within the solenoid compared to outside. The video illustrates setting up a rectangular loop around the solenoid to ascertain the internal field strength. The derivation culminates in the equation B = μ₀nI, explaining the relationships between current, loop density, and field strength, stressing the enhancements gained by more turns and higher current.

  • 00:50:00 - 00:55:00

    Further explanations on solenoids involve solving an example problem to compute the magnetic field strength based on given parameters like the number of turns, current, and solenoid length. The problem-solving process illustrates how changes in solenoid design impact field strength, reinforcing the conceptual explanations with quantitative analysis. This part emphasizes optimizing design for maximum desired magnetic field by manipulating variables such as solenoid length and wire turns.

  • 00:55:00 - 01:00:00

    The video delves into torque experienced by current loops in magnetic fields, showing how forces at different sections and angles cause rotational effects. The arising torques are analyzed with respect to magnetic field direction, showcasing when maximum torque occurs. Derivations lead to a formula for torque considering the number of coils, current, area, and magnetic field orientation. Practical examples illustrate calculating maximum torque aligning with sections of the coil in the magnetic field, solving exemplar problems to solidify understanding.

  • 01:00:00 - 01:05:00

    The interpretation of torque on current-carrying loops involves attention to geometric orientation concerning magnetic fields. By viewing from different perspectives (top view and side view), the video illustrates conditions where torque is maximized, diminished, or nullified based on alignment. Important outcomes are noted, such as equilibrium positions with zero net torque. The role of orientation changes in affecting loop dynamics is clarified, enriching comprehension through visual scenarios and practical computation of torques.

  • 01:05:00 - 01:10:00

    A specific problem is presented involving calculating torque exerted on a circular coil in a magnetic field, using known elements such as radius, number of loops, current, and field strength. How these variables influence torque is demonstrated, stressing the impact of the loop area. Viewers are walked through calculating torque using predefined equations, reinforcing the grasp on determining maximum scenarios within magnetic interactions.

  • 01:10:00 - 01:15:00

    Explaining an additional problem on magnetic fields and torque, the video traverses calculating necessary magnetic field strength to produce a known torque from a coil. This involves rearranging the torque equation, substituting given quantities, and deriving the magnetic field value. This section solidifies understanding by blending qualitative comprehension with numeric problem-solving, effectively tackling field-related quandaries using theoretical frameworks established earlier.

  • 01:15:00 - 01:22:40

    Finally, a concluding remark emphasizes wherever/whenever such videos delve into enriching educational content, more available content for learning physics is pitched, encouraging viewers into engaging further with educational resources provided, thanking them for their continuation in such exploration of electromagnetic principles.

Afficher plus

Carte mentale

Mind Map

Questions fréquemment posées

  • What happens when the north pole of a magnet faces the north pole of another?

    The magnets repel each other.

  • How are magnetic fields created?

    Magnetic fields are created by moving electric charges.

  • What is the right-hand rule?

    The right-hand rule is a way to determine the direction of the magnetic field in relation to the direction of the current.

  • What is the formula for calculating the strength of a magnetic field near a wire?

    B = μ₀ * I / 2πr, where B is the magnetic field, I is the current, and r is the distance from the wire.

  • What affects the magnitude of the magnetic force on a wire?

    The magnetic force is affected by the current strength, the magnetic field strength, and the length of the wire.

  • When is the magnetic force on a moving charge at its maximum?

    When the velocity and the magnetic field are perpendicular, the magnetic force is at its maximum.

  • What happens to the magnetic field strength as distance from the wire increases?

    The magnetic field strength decreases as the distance from the wire increases.

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Sous-titres
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Défilement automatique:
  • 00:00:01
    in this video we're going to talk about
  • 00:00:02
    magnetism
  • 00:00:05
    perhaps you're familiar with bar magnets
  • 00:00:09
    and you know that
  • 00:00:11
    when you place the north pole
  • 00:00:14
    of a barge magnet with another north
  • 00:00:16
    pole
  • 00:00:17
    these two they repel
  • 00:00:20
    they're gonna push apart
  • 00:00:22
    however
  • 00:00:24
    let's say if
  • 00:00:26
    you were to face
  • 00:00:28
    the north pole of one bar magnet
  • 00:00:30
    with the south pole of another
  • 00:00:35
    they won't repel these two will feel
  • 00:00:37
    a force of attraction
  • 00:00:43
    the north floor is attracted to the
  • 00:00:44
    south pole
  • 00:00:46
    but if you put the north pole with
  • 00:00:47
    another north pole of another magnet
  • 00:00:49
    they will repel or if you put the south
  • 00:00:51
    pole of one magnet with the south for
  • 00:00:53
    another they will also repel
  • 00:00:57
    every barbed magnet has its own magnetic
  • 00:01:00
    field
  • 00:01:01
    which emanates away from the north pole
  • 00:01:04
    and it travels towards the south pole
  • 00:01:10
    likewise this one has
  • 00:01:13
    a magnetic field emanating away from the
  • 00:01:15
    north pole notice that the magnetic
  • 00:01:17
    field cancels in the middle
  • 00:01:21
    those two they repel each other
  • 00:01:23
    and in the case of the other example
  • 00:01:28
    the magnetic field leaves the north pole
  • 00:01:33
    but it enters the south pole
  • 00:01:40
    so in the case of attraction notice that
  • 00:01:42
    the magnetic field
  • 00:01:44
    between the north and south pole of the
  • 00:01:47
    two barmatics notice that they're in the
  • 00:01:49
    same direction
  • 00:01:51
    and so the additive you have that force
  • 00:01:52
    of attraction
  • 00:01:56
    so what causes magnetic fields
  • 00:01:59
    magnetic fields are created by moving
  • 00:02:02
    electric charge
  • 00:02:04
    and this example can be illustrated if
  • 00:02:06
    you have a wire
  • 00:02:10
    whenever there's an electric current
  • 00:02:12
    that flows through a wire
  • 00:02:15
    it creates its own magnetic field
  • 00:02:19
    and the magnetic field created by this
  • 00:02:22
    wire it looks like this
  • 00:02:25
    it's circular
  • 00:02:30
    on the left side
  • 00:02:32
    the magnetic field is leaving the page
  • 00:02:34
    so it's represented by
  • 00:02:37
    a dot or circle
  • 00:02:39
    and on the right side it enters the page
  • 00:02:42
    which is symbolized by an x you need to
  • 00:02:44
    be familiar with that by the way
  • 00:02:47
    you can use the right hand rule to
  • 00:02:49
    figure this out
  • 00:02:50
    if you take your hand
  • 00:02:52
    and if you
  • 00:02:54
    curl it around a pen
  • 00:02:56
    with your thumb facing
  • 00:02:58
    the direction of the current
  • 00:03:00
    the way your fingers curl around the pen
  • 00:03:02
    is the way the magnetic field travels
  • 00:03:04
    around
  • 00:03:05
    the wire
  • 00:03:08
    try that
  • 00:03:10
    so here's a picture that describes it
  • 00:03:12
    as you can see
  • 00:03:14
    you want the thumb facing the direction
  • 00:03:16
    of the current
  • 00:03:19
    which is upward and as you curl your
  • 00:03:21
    hand
  • 00:03:22
    around the wire
  • 00:03:25
    notice the way your hand curls around it
  • 00:03:28
    on the left side
  • 00:03:30
    you can see how it's
  • 00:03:33
    coming out of the page
  • 00:03:35
    and then on the right side
  • 00:03:37
    your hand curls into the page
  • 00:03:40
    and so the way your hand curls around
  • 00:03:42
    that wire is the way the magnetic field
  • 00:03:45
    created by the moving charge
  • 00:03:48
    travels around the wire it's out of the
  • 00:03:50
    page on the left side
  • 00:03:53
    and it's going into the page on the
  • 00:03:55
    right side
  • 00:03:59
    now there's an equation that allows you
  • 00:04:00
    to calculate the strength of the
  • 00:04:02
    magnetic field created by such a wire
  • 00:04:05
    and here's the equation
  • 00:04:07
    b is equal to u0
  • 00:04:09
    times i
  • 00:04:11
    divided by 2 pi r
  • 00:04:15
    so let's say if you want to calculate
  • 00:04:17
    the magnetic field
  • 00:04:19
    some distance point a
  • 00:04:21
    away from the wire
  • 00:04:23
    r is the distance between the wire
  • 00:04:26
    and point a
  • 00:04:28
    b is the strength of the magnetic field
  • 00:04:32
    and b is measured in units of tesla or
  • 00:04:36
    capital t
  • 00:04:39
    u 0 or mu 0 it's equal to four pi
  • 00:04:43
    times ten to the minus seven
  • 00:04:46
    this is known as the permeability of
  • 00:04:49
    free space
  • 00:04:50
    and the units are tesla
  • 00:04:52
    times meters per
  • 00:04:54
    amp
  • 00:04:56
    now notice that the current
  • 00:04:58
    and the magnetic field are directly
  • 00:05:00
    related
  • 00:05:01
    if you increase the magnitude of the
  • 00:05:03
    current
  • 00:05:04
    the strength of the magnetic field
  • 00:05:06
    generated by this wire will increase as
  • 00:05:08
    well
  • 00:05:10
    and the reason for that is because the
  • 00:05:12
    current is on the top of the fraction
  • 00:05:15
    whenever you increase the numerator of a
  • 00:05:17
    fraction the value of the entire
  • 00:05:19
    fraction will increase
  • 00:05:21
    now r is on the bottom
  • 00:05:23
    so that means that r is inversely
  • 00:05:25
    related to b
  • 00:05:27
    if you increase
  • 00:05:28
    the distance between the wire and the
  • 00:05:31
    magnetic and uh point of interest i
  • 00:05:34
    should say
  • 00:05:35
    the magnetic field at that point will be
  • 00:05:38
    weaker
  • 00:05:40
    as you move away from the wire the
  • 00:05:42
    strength of the magnetic field weakens
  • 00:05:46
    by the way the number of magnetic lines
  • 00:05:47
    that you see in a picture is
  • 00:05:49
    proportional to the strength of the
  • 00:05:51
    magnetic field
  • 00:05:55
    so for example
  • 00:05:57
    let's say if the magnetic field
  • 00:05:59
    in this region looks like this
  • 00:06:02
    versus
  • 00:06:04
    two lines as opposed to three
  • 00:06:07
    the magnetic field on the left is
  • 00:06:08
    stronger than the one on the right
  • 00:06:11
    if you have more lines that are closer
  • 00:06:13
    together the strength of the magnetic
  • 00:06:15
    field is stronger
  • 00:06:16
    but make sure you know this relationship
  • 00:06:18
    so anytime you increase the electric
  • 00:06:20
    current in a wire
  • 00:06:21
    the strength of the magnetic field will
  • 00:06:23
    increase
  • 00:06:25
    and as you move away from the wire
  • 00:06:27
    the strength of the magnetic field will
  • 00:06:29
    decrease
  • 00:06:31
    let's work on this problem a vertical
  • 00:06:33
    wire carries a current of 45 amps do
  • 00:06:36
    self
  • 00:06:38
    calculate the magnitude
  • 00:06:39
    and the direction of the magnetic field
  • 00:06:41
    two centimeters to the right of the wire
  • 00:06:44
    so go ahead and try this problem
  • 00:06:46
    so let's say this
  • 00:06:48
    is the wire
  • 00:06:51
    and the current is due south
  • 00:06:54
    so it's going down
  • 00:06:59
    and that means that using the right hand
  • 00:07:01
    rule the magnetic field is going to
  • 00:07:04
    enter the page
  • 00:07:05
    on the right side
  • 00:07:07
    and it's going to be out of the page
  • 00:07:10
    i mean it enters the page on the left
  • 00:07:12
    side
  • 00:07:13
    but on the right side it comes out of
  • 00:07:15
    the page
  • 00:07:17
    so when it enters the page put it x
  • 00:07:20
    and when it leaves the page
  • 00:07:22
    put a circle a closed circle
  • 00:07:27
    now our goal is to find the magnetic
  • 00:07:28
    field two centimeters
  • 00:07:30
    to the right of the wire
  • 00:07:32
    so we already have the direction of the
  • 00:07:34
    magnetic field
  • 00:07:36
    it's out of the page
  • 00:07:41
    all we need to do now is calculate the
  • 00:07:42
    magnitude
  • 00:07:44
    so we can use this equation
  • 00:07:46
    b is equal to mu zero
  • 00:07:49
    times i
  • 00:07:50
    divided by two pi r
  • 00:07:54
    mu zero is equal to four pi
  • 00:07:58
    times 10 to the minus 7
  • 00:08:01
    and the current
  • 00:08:02
    is 45 amps
  • 00:08:08
    r is the distance between the wire and
  • 00:08:10
    the point of interest
  • 00:08:12
    in this case r is two centimeters but we
  • 00:08:14
    need to convert that to meters
  • 00:08:16
    so we got to divide by a hundred
  • 00:08:19
    one meter is equal to 100 centimeters
  • 00:08:23
    so that's going to be point .02
  • 00:08:26
    meters
  • 00:08:28
    so all you got to do is just type this
  • 00:08:30
    in
  • 00:08:45
    and you should get
  • 00:08:53
    4.5
  • 00:08:55
    times 10 to the negative 4
  • 00:08:57
    tesla
  • 00:08:59
    so that's going to be the strength of
  • 00:09:00
    the magnetic field
  • 00:09:01
    2 centimeters away from the wire
  • 00:09:04
    number 2
  • 00:09:06
    a wire carries a current of 10 amps
  • 00:09:09
    at what distance from the wire
  • 00:09:12
    will a magnetic field of
  • 00:09:14
    8 times 10 to the minus 4 teslas be
  • 00:09:16
    produced
  • 00:09:19
    so we got to find r in this case
  • 00:09:21
    we can use the same equation
  • 00:09:24
    we don't have to worry about the
  • 00:09:25
    direction so we don't really need to
  • 00:09:27
    draw a picture
  • 00:09:29
    so let's solve for r
  • 00:09:31
    let's multiply both sides by r
  • 00:09:34
    so on the right side it's going to
  • 00:09:36
    cancel
  • 00:09:39
    so b times r
  • 00:09:41
    is equal to mu 0
  • 00:09:42
    times i divided by 2 pi
  • 00:09:45
    now let's multiply both sides of the
  • 00:09:47
    equation by 1 over b
  • 00:09:51
    if we do that on the left side b will
  • 00:09:53
    cancel
  • 00:09:56
    so the distance is going to be mu zero
  • 00:09:59
    times i
  • 00:10:01
    divided by
  • 00:10:02
    two pi
  • 00:10:03
    times b
  • 00:10:04
    so it's 4 pi
  • 00:10:06
    times 10 to the minus 7
  • 00:10:09
    times the current of 10 amps
  • 00:10:14
    divided by 2 pi
  • 00:10:17
    times the strength of the magnetic field
  • 00:10:22
    so we can cancel pi
  • 00:10:24
    in fact four pi divided by two pi is
  • 00:10:27
    just two
  • 00:10:28
    so it's going to be two times ten to the
  • 00:10:30
    minus seven times ten
  • 00:10:33
    divided by eight times ten to minus four
  • 00:10:39
    and so this is equal to two point five
  • 00:10:43
    times ten to the minus three
  • 00:10:45
    and the units is meters
  • 00:10:48
    so if we want to we can convert it to
  • 00:10:50
    millimeters
  • 00:10:53
    and to do that
  • 00:10:56
    you need to divide actually multiply by
  • 00:10:59
    a thousand there's a thousand
  • 00:11:00
    millimeters per meter
  • 00:11:05
    if you multiply by a thousand
  • 00:11:07
    this will give you 2.5
  • 00:11:10
    millimeters
  • 00:11:12
    and so that's the answer that's how far
  • 00:11:14
    away from the wire
  • 00:11:16
    that
  • 00:11:17
    you have to be to measure a magnetic
  • 00:11:19
    field of eight times ten to minus four
  • 00:11:24
    by the way if you ever were to place a
  • 00:11:26
    compass near a wire
  • 00:11:29
    whenever there's an electric current
  • 00:11:30
    flowing through that wire
  • 00:11:32
    it will cause the compass to deflect you
  • 00:11:34
    should try it
  • 00:11:40
    now let's say if we have
  • 00:11:42
    a current carrion wire
  • 00:11:45
    what's going to happen if we place this
  • 00:11:47
    wire
  • 00:11:48
    inside a magnetic field let's say the
  • 00:11:50
    magnetic field
  • 00:11:51
    is directed east
  • 00:11:54
    and the current is moving north
  • 00:11:58
    a magnetic field
  • 00:12:00
    exerts no force on a stationary charge
  • 00:12:04
    however if the electric charge is moving
  • 00:12:07
    then the magnetic field will exert a
  • 00:12:10
    force specifically a magnetic force
  • 00:12:13
    so whenever you have a wire
  • 00:12:15
    with an electric current that means you
  • 00:12:18
    have moving charges in the wire
  • 00:12:20
    the magnetic field will exert a force
  • 00:12:23
    on the wire
  • 00:12:25
    you can calculate the strength
  • 00:12:27
    of the magnetic force using this
  • 00:12:29
    equation
  • 00:12:30
    f is equal to i
  • 00:12:32
    lb
  • 00:12:34
    sine theta
  • 00:12:36
    so the strength of the magnetic force is
  • 00:12:38
    proportional to the current
  • 00:12:40
    if you increase the current
  • 00:12:42
    the magnetic force
  • 00:12:44
    will
  • 00:12:45
    increase
  • 00:12:50
    the magnetic force is also proportional
  • 00:12:53
    to the strength of the magnetic field
  • 00:12:56
    if you increase the magnetic field the
  • 00:12:58
    magnetic force will increase as well
  • 00:13:00
    and it's also proportional to the length
  • 00:13:02
    of the wire
  • 00:13:06
    now depends on the angle as well
  • 00:13:11
    so here's one example where the current
  • 00:13:13
    and the magnetic field are perpendicular
  • 00:13:16
    and here's another example when
  • 00:13:18
    the current and the magnetic field are
  • 00:13:20
    at an angle
  • 00:13:22
    and here's one that's parallel
  • 00:13:26
    now
  • 00:13:27
    theta
  • 00:13:28
    is the angle between the current and the
  • 00:13:30
    magnetic field
  • 00:13:32
    when they are perpendicular sine 90 is
  • 00:13:35
    equal to one
  • 00:13:37
    and one basically represents a hundred
  • 00:13:39
    percent
  • 00:13:40
    so the maximum force occurs when
  • 00:13:43
    the current and the magnetic field are
  • 00:13:45
    perpendicular to each other
  • 00:13:51
    now when it's at an angle
  • 00:13:54
    it's going to be between anywhere from
  • 00:13:56
    zero to 100 percent of its maximum value
  • 00:14:01
    so then you can use this equation when
  • 00:14:03
    they're parallel
  • 00:14:04
    the angle is equal to zero degrees
  • 00:14:07
    sine zero
  • 00:14:09
    is equal to zero therefore
  • 00:14:12
    the magnetic field exerts no magnetic
  • 00:14:14
    force
  • 00:14:16
    on a moving charge that moves parallel
  • 00:14:19
    or even anti-parallel to magnetic field
  • 00:14:25
    so for the third example there's no
  • 00:14:27
    magnetic force
  • 00:14:28
    acting on the current
  • 00:14:32
    they have to be
  • 00:14:33
    at an angle with respect to each other
  • 00:14:36
    they can be parallel
  • 00:14:38
    now what about the direction
  • 00:14:41
    let's go back to our last example so
  • 00:14:44
    let's say if the current
  • 00:14:46
    is due north
  • 00:14:48
    and magnetic field
  • 00:14:50
    is directed east
  • 00:14:53
    in what direction is the
  • 00:14:56
    force now the force has to be
  • 00:14:59
    perpendicular
  • 00:15:00
    to the current anti-magnetic field
  • 00:15:04
    so if the current is in
  • 00:15:06
    the north-south direction
  • 00:15:08
    and if the magnetic field is in the west
  • 00:15:10
    east direction
  • 00:15:12
    then the force is either into the page
  • 00:15:14
    or out of the page
  • 00:15:16
    that is along the z axis
  • 00:15:19
    so how can we figure this out
  • 00:15:21
    well we have to use
  • 00:15:22
    the right hand rule
  • 00:15:24
    so take your right hand
  • 00:15:26
    and you want to extend it
  • 00:15:30
    you want
  • 00:15:32
    your thumb
  • 00:15:33
    to be in the direction of the current
  • 00:15:36
    and you want your other four fingers
  • 00:15:39
    to be in the direction of the magnetic
  • 00:15:41
    field
  • 00:15:44
    so this represents b
  • 00:15:46
    and this represents the current i
  • 00:15:49
    so using your right hand look at where
  • 00:15:51
    your right hand opens towards
  • 00:15:53
    if you direct it the way it's presented
  • 00:15:55
    here
  • 00:15:56
    it's going to go into the page
  • 00:15:59
    the force comes out of the palm of your
  • 00:16:01
    right hand
  • 00:16:02
    and so it's going to be directed uh
  • 00:16:06
    into the
  • 00:16:06
    page
  • 00:16:08
    and that's how you can figure it out
  • 00:16:10
    let's try another example
  • 00:16:12
    so let's say if we have a wire
  • 00:16:16
    and a current is directed east
  • 00:16:20
    and the magnetic field
  • 00:16:22
    is directed into the page
  • 00:16:26
    and what direction is the force going to
  • 00:16:28
    be
  • 00:16:29
    so if the current is in the east-west
  • 00:16:31
    direction
  • 00:16:32
    and
  • 00:16:33
    well the current's really west to east
  • 00:16:36
    and the magnetic field is in
  • 00:16:38
    the z direction that is between out of
  • 00:16:40
    the page and into the page
  • 00:16:42
    then the force has to be in a north
  • 00:16:44
    south direction
  • 00:16:46
    these three
  • 00:16:48
    variables have to be perpendicular to
  • 00:16:49
    each other
  • 00:16:52
    so what you want to do this time
  • 00:16:55
    you want to direct your four fingers
  • 00:16:58
    into the page
  • 00:17:02
    and you want your thumb
  • 00:17:05
    directed
  • 00:17:07
    east that is in the direction of the
  • 00:17:09
    current
  • 00:17:13
    but you want your four fingers to be
  • 00:17:14
    into the page
  • 00:17:16
    and the force comes out of the palm of
  • 00:17:18
    your right hand
  • 00:17:20
    so the force will be directed
  • 00:17:22
    north
  • 00:17:23
    if you do that
  • 00:17:25
    it's kind of hard to draw
  • 00:17:27
    the direction of the hand
  • 00:17:29
    on this video but hopefully you can
  • 00:17:32
    visualize it
  • 00:17:36
    number three
  • 00:17:37
    a 2.5 meter long wire
  • 00:17:40
    carries a current of 5 amps in the
  • 00:17:42
    presence
  • 00:17:43
    of a magnetic field with a strength of 2
  • 00:17:46
    x 10-3 teslas
  • 00:17:49
    calculate the magnitude of the magnetic
  • 00:17:51
    force on the wire
  • 00:17:53
    using the picture shown below
  • 00:17:55
    so
  • 00:17:56
    feel free to try that
  • 00:17:58
    the equation that we need is f is equal
  • 00:18:00
    to i
  • 00:18:01
    lb
  • 00:18:03
    sine theta
  • 00:18:04
    now let's talk about the angle
  • 00:18:06
    so the current is directed east
  • 00:18:09
    and the magnetic field
  • 00:18:11
    is directed 30 degrees
  • 00:18:13
    relative to the horizontal
  • 00:18:16
    so theta is always going to be the angle
  • 00:18:18
    between
  • 00:18:19
    the magnetic field
  • 00:18:21
    and the current
  • 00:18:24
    so you can also use this angle as well
  • 00:18:26
    because
  • 00:18:27
    that angle is between b and i
  • 00:18:31
    180 minus 30 is 150 and it turns out
  • 00:18:34
    that sine of 150 and sine of 30
  • 00:18:38
    they're both equal to one half
  • 00:18:40
    so it doesn't matter
  • 00:18:41
    so whether you choose this angle which
  • 00:18:43
    is between
  • 00:18:44
    i and b or if you use this angle
  • 00:18:47
    the answer will be the same
  • 00:18:50
    so just something to know
  • 00:18:52
    so now let's go ahead and calculate f
  • 00:18:54
    so it's going to be the current which is
  • 00:18:56
    5
  • 00:18:57
    times
  • 00:18:58
    the length of the wire
  • 00:19:01
    which is uh 2.5 meters
  • 00:19:04
    times the strength for the magnetic
  • 00:19:05
    field which is uh
  • 00:19:07
    2 times 10 to the minus 3. multiplied by
  • 00:19:11
    sine
  • 00:19:12
    of thirty
  • 00:19:23
    so the magnetic force
  • 00:19:25
    is very small it's
  • 00:19:27
    .0125 newtons
  • 00:19:31
    and so that's going to be the force
  • 00:19:32
    exerted on this current carrying wire
  • 00:19:38
    number four
  • 00:19:40
    a current of 35 amps
  • 00:19:42
    flows due west in a wire that
  • 00:19:44
    experiences a magnetic force
  • 00:19:46
    of 0.75 newtons per meter
  • 00:19:50
    what is the strength of the magnetic
  • 00:19:51
    field which is directed to south
  • 00:19:56
    so here's the wire
  • 00:19:58
    and the current
  • 00:19:59
    is directed
  • 00:20:01
    west
  • 00:20:02
    and the magnetic field
  • 00:20:04
    is directed itself
  • 00:20:10
    our goal is to find the strength of the
  • 00:20:11
    magnetic field we need to solve for b
  • 00:20:14
    so let's write the equation
  • 00:20:16
    f is equal to ilb
  • 00:20:19
    now because the current and the magnetic
  • 00:20:21
    field
  • 00:20:22
    are at right angles
  • 00:20:24
    to each other because it's 90 degrees
  • 00:20:26
    sine 90 is one
  • 00:20:28
    so we don't need the sine portion of
  • 00:20:29
    this equation
  • 00:20:31
    now we're given the force
  • 00:20:33
    per meter
  • 00:20:35
    so that's f divided by l
  • 00:20:38
    if we divide both sides by l we're going
  • 00:20:40
    to get this equation f over l
  • 00:20:43
    is equal to
  • 00:20:45
    the current multiplied by the magnetic
  • 00:20:47
    field
  • 00:20:48
    we have the force per meter
  • 00:20:50
    that's 0.75
  • 00:20:55
    that value
  • 00:20:57
    takes care of two of these variables
  • 00:21:00
    we have the current which is 35 amps so
  • 00:21:03
    we got to solve for b
  • 00:21:07
    so it's just going to be 0.75
  • 00:21:10
    divided by 35
  • 00:21:12
    and so b
  • 00:21:14
    is equal to
  • 00:21:16
    0.0214 tesla
  • 00:21:20
    so anytime you have the force per unit
  • 00:21:22
    left
  • 00:21:23
    or
  • 00:21:24
    newtons per meter
  • 00:21:27
    make sure you understand that it's f
  • 00:21:29
    divided by l the entire thing is f over
  • 00:21:31
    l
  • 00:21:33
    so you might see that expression in this
  • 00:21:35
    chapter
  • 00:21:36
    in a few problems
  • 00:21:38
    now what is the direction
  • 00:21:41
    of the magnetic force
  • 00:21:45
    we know the current flows west the
  • 00:21:48
    magnetic field is south
  • 00:21:50
    so the force is either
  • 00:21:52
    into the page or out of the page
  • 00:21:56
    so what you want to do
  • 00:21:58
    is you want to direct
  • 00:22:00
    your four fingers
  • 00:22:02
    south
  • 00:22:04
    and your thumb
  • 00:22:05
    west
  • 00:22:07
    and the force should come out of the
  • 00:22:09
    page
  • 00:22:14
    so let's see if i can draw that
  • 00:22:17
    so you want your thumb facing this way
  • 00:22:21
    and the four before fingers
  • 00:22:23
    facing this way
  • 00:22:25
    using your right hand if you do that
  • 00:22:32
    i put into the page this should be out
  • 00:22:34
    of the page
  • 00:22:39
    the force should come out
  • 00:22:41
    of the palm of your hand
  • 00:22:43
    so make sure the current is aligned with
  • 00:22:45
    your thumb
  • 00:22:46
    the magnetic field is aligned with
  • 00:22:50
    your four fingers
  • 00:22:52
    and then the force should come out of
  • 00:22:53
    the palm behind that's out of the page
  • 00:22:57
    now let's say if we have
  • 00:22:59
    a rectangular
  • 00:23:01
    metal loop
  • 00:23:05
    with a current
  • 00:23:06
    that flows in the metal loop clockwise
  • 00:23:10
    so in this section the current's going
  • 00:23:12
    up
  • 00:23:13
    here it's going down
  • 00:23:15
    here it's directed towards the right and
  • 00:23:16
    here is directed towards left
  • 00:23:18
    and only a portion of this loop
  • 00:23:21
    is inside
  • 00:23:23
    a magnetic field that is only the
  • 00:23:26
    bottom portion
  • 00:23:32
    now let's say the magnetic field
  • 00:23:34
    is directed
  • 00:23:36
    into the page
  • 00:23:38
    so i'm going to put x everywhere
  • 00:23:47
    where will
  • 00:23:49
    this rectangular loop move if initially
  • 00:23:53
    it's at rest will it move towards the
  • 00:23:54
    right will it begin moving towards the
  • 00:23:56
    left
  • 00:23:58
    up or down what would you say
  • 00:24:01
    so let's start with this portion of the
  • 00:24:04
    wire
  • 00:24:05
    or of the the metal loop
  • 00:24:07
    let's see what the magnetic force
  • 00:24:09
    on that portion
  • 00:24:11
    is directed
  • 00:24:15
    so what you want to do is you want to
  • 00:24:18
    place your thumb
  • 00:24:20
    facing south
  • 00:24:22
    you want the four fingers of your hand
  • 00:24:23
    to be going into the page
  • 00:24:25
    and the magnetic force
  • 00:24:27
    should be directed
  • 00:24:29
    east
  • 00:24:30
    so let's see if i can draw that
  • 00:24:38
    so you want your four fingers
  • 00:24:40
    going into the page
  • 00:24:42
    you want your thumb
  • 00:24:43
    going south
  • 00:24:51
    so
  • 00:24:52
    make sure the magnetic field or your
  • 00:24:54
    four fingers is going into the page
  • 00:24:57
    and your thumb is in the direction of
  • 00:24:59
    the current
  • 00:25:03
    and the force should come out of your
  • 00:25:04
    hand
  • 00:25:06
    and that is out of the palm of your
  • 00:25:08
    right hand
  • 00:25:09
    and it should be directed east if you do
  • 00:25:11
    it correctly
  • 00:25:13
    so make sure you try that and make sure
  • 00:25:15
    you can master this right hand rule
  • 00:25:20
    now
  • 00:25:22
    for the other side the left side
  • 00:25:24
    everything is the same except the
  • 00:25:25
    current
  • 00:25:26
    because the current is in the opposite
  • 00:25:28
    direction
  • 00:25:29
    the force has to be in the opposite
  • 00:25:30
    direction
  • 00:25:33
    now these two forces are equal in
  • 00:25:35
    magnitude and because they're opposite
  • 00:25:37
    direction
  • 00:25:38
    they will cancel out
  • 00:25:40
    so the loop is not going to move towards
  • 00:25:42
    the left or towards the right these
  • 00:25:44
    forces balance each other out
  • 00:25:47
    now in the top part of the loop
  • 00:25:50
    there's no magnetic field in that region
  • 00:25:52
    so therefore there's no force
  • 00:25:56
    the net force
  • 00:25:57
    is going to be based on
  • 00:25:59
    this portion of the loop
  • 00:26:03
    because it's not balanced by
  • 00:26:05
    this portion of the loop if the entire
  • 00:26:07
    loop was in the magnetic field
  • 00:26:09
    all the forces will cancel
  • 00:26:11
    but since it's not
  • 00:26:14
    this one will create a net force
  • 00:26:17
    so now let's draw another picture point
  • 00:26:19
    your thumb towards the left and make
  • 00:26:21
    sure your four fingers are going into
  • 00:26:23
    the page if you do it correctly
  • 00:26:26
    the force
  • 00:26:27
    should be direct itself
  • 00:26:32
    so what you want to do is you want your
  • 00:26:34
    thumb directed uh
  • 00:26:36
    west
  • 00:26:38
    and you want your four fingers
  • 00:26:41
    going into the page
  • 00:26:53
    so if you do it correctly
  • 00:26:54
    the magnetic force should be coming
  • 00:26:56
    south
  • 00:26:57
    out
  • 00:26:58
    of the palm of your right hand
  • 00:27:03
    now let's move on to another topic we
  • 00:27:06
    talked about how to calculate the
  • 00:27:07
    magnetic force
  • 00:27:09
    on a current carrying wire
  • 00:27:11
    but what about the magnetic force on a
  • 00:27:13
    single point charge
  • 00:27:15
    because any move in charge
  • 00:27:17
    will create a magnetic field so let's go
  • 00:27:20
    ahead and come up with an equation
  • 00:27:22
    let's start with this equation f is
  • 00:27:24
    equal to ilb
  • 00:27:26
    sine beta
  • 00:27:30
    current
  • 00:27:32
    is defined as the amount of electric
  • 00:27:34
    charge that flows
  • 00:27:35
    per unit time
  • 00:27:38
    and capital q is the total amount of
  • 00:27:40
    charge it can be due to many charged
  • 00:27:42
    particles
  • 00:27:44
    so capital q is going to be equal to
  • 00:27:46
    lower lowercase q which is the magnitude
  • 00:27:48
    of each
  • 00:27:50
    charged particle
  • 00:27:51
    times n which is the number of charged
  • 00:27:53
    particles
  • 00:27:55
    and that will give you the total charge
  • 00:27:57
    divided by t
  • 00:28:01
    now distance is equal to the speed
  • 00:28:04
    multiplied by the time
  • 00:28:07
    and length
  • 00:28:09
    could be thought of as distance they're
  • 00:28:10
    both measured in meters
  • 00:28:12
    so we can replace out with vt so let's
  • 00:28:16
    replace i
  • 00:28:19
    with q n over t
  • 00:28:26
    and let's replace
  • 00:28:28
    l with v t
  • 00:28:38
    so we can cancel t
  • 00:28:41
    now if you want to find the magnetic
  • 00:28:42
    force
  • 00:28:43
    on a single point charge
  • 00:28:46
    that means there's only one charged
  • 00:28:48
    particle so n is one
  • 00:28:50
    so when n is one we can get rid of it
  • 00:28:53
    so this leaves us with f is equal to
  • 00:28:57
    bqv
  • 00:28:58
    sine theta
  • 00:29:00
    so that's how you can calculate the
  • 00:29:01
    magnetic force
  • 00:29:03
    on a moving charge particle when it's
  • 00:29:05
    inside a magnetic field
  • 00:29:09
    so let's say if we have a proton
  • 00:29:13
    and let's say it's moving towards the
  • 00:29:15
    right
  • 00:29:16
    and also the magnetic field is directed
  • 00:29:18
    towards the right
  • 00:29:20
    if these two are parallel
  • 00:29:23
    sine zero is equal to zero so there's
  • 00:29:26
    going to be no magnetic force
  • 00:29:30
    they have to be perpendicular
  • 00:29:34
    now if the proton
  • 00:29:36
    is moving at an angle
  • 00:29:39
    relative to the magnetic field
  • 00:29:43
    and then you can use the equation f is
  • 00:29:45
    equal to b q v
  • 00:29:47
    sine theta
  • 00:29:49
    where theta is the angle between
  • 00:29:51
    the magnetic field and the velocity
  • 00:29:53
    vector
  • 00:29:57
    now let's say if the proton
  • 00:29:59
    it's moving perpendicular
  • 00:30:01
    to the magnetic field
  • 00:30:04
    that is they're at right angles or 90
  • 00:30:06
    degrees relative to each other
  • 00:30:08
    sine 90
  • 00:30:10
    is one
  • 00:30:12
    so the magnetic force
  • 00:30:14
    will have its maximum value at this
  • 00:30:16
    point and it's equal to simply bqv
  • 00:30:23
    now let's say if
  • 00:30:26
    we have a proton
  • 00:30:28
    and it's moving towards the right
  • 00:30:33
    and the magnetic force or rather the
  • 00:30:36
    magnetic field
  • 00:30:37
    is directed north
  • 00:30:39
    what is the direction of the magnetic
  • 00:30:42
    force
  • 00:30:45
    it has to be in the z direction if the
  • 00:30:47
    velocity is in the x direction and if
  • 00:30:50
    the magnetic field is in the y direction
  • 00:30:52
    the magnetic force have to be in the z
  • 00:30:54
    direction
  • 00:30:55
    and you can use the right hand rule
  • 00:30:58
    think of velocity as the current so to
  • 00:31:00
    speak that's where the charged particles
  • 00:31:01
    are moving
  • 00:31:04
    so you can use the
  • 00:31:06
    the same right hand rule as we've been
  • 00:31:07
    doing
  • 00:31:08
    you want your four fingers
  • 00:31:10
    to be in the direction of magnetic field
  • 00:31:13
    and you want your thumb
  • 00:31:16
    to be
  • 00:31:17
    in a direction of the velocity
  • 00:31:26
    so if you orient your right hand
  • 00:31:28
    this way
  • 00:31:30
    the force should come out of the palm of
  • 00:31:32
    your right hand
  • 00:31:33
    and so
  • 00:31:34
    it should come
  • 00:31:35
    out of the page
  • 00:31:41
    so i'm going to put a circle for that
  • 00:31:46
    now
  • 00:31:47
    for a proton the magnetic force is out
  • 00:31:50
    of the page
  • 00:31:51
    but
  • 00:31:52
    for an electron
  • 00:31:54
    the magnetic force will be in the
  • 00:31:57
    opposite direction
  • 00:31:58
    that is
  • 00:31:59
    it's going to be going
  • 00:32:01
    into the page
  • 00:32:03
    so for any negatively charged particle
  • 00:32:06
    simply reverse the direction of magnetic
  • 00:32:08
    force if you need to find it
  • 00:32:12
    for a positively charged particle
  • 00:32:14
    it's going to work in the exact same way
  • 00:32:17
    as the right hand rule directs it
  • 00:32:19
    number five
  • 00:32:20
    a proton moves east with a speed of four
  • 00:32:23
    times ten to the sixth
  • 00:32:25
    meters per second
  • 00:32:27
    in a magnetic field of two times ten to
  • 00:32:29
    minus four teslas
  • 00:32:31
    directed into the page
  • 00:32:34
    what is the magnitude of the magnetic
  • 00:32:36
    force acting on a proton
  • 00:32:40
    so if we need to find the magnitude we
  • 00:32:41
    don't have to worry about the direction
  • 00:32:43
    all we need to know is that
  • 00:32:45
    the velocity
  • 00:32:48
    and
  • 00:32:49
    the magnetic field are perpendicular to
  • 00:32:51
    each other
  • 00:32:53
    the proton is moving east in the x
  • 00:32:55
    direction
  • 00:32:56
    the magnetic field is directed into the
  • 00:32:58
    page
  • 00:33:00
    that is in a negative z direction
  • 00:33:03
    so therefore the magnetic force
  • 00:33:07
    has to be in the y direction
  • 00:33:13
    so all we need to do is find the
  • 00:33:14
    magnitude
  • 00:33:15
    so we just got to use the equation f is
  • 00:33:18
    equal to b q v
  • 00:33:19
    sine theta
  • 00:33:21
    now because the velocity and the
  • 00:33:24
    magnetic field are perpendicular
  • 00:33:26
    that is the velocities in the x
  • 00:33:28
    direction the magnetic field is in the z
  • 00:33:30
    direction
  • 00:33:31
    the angle has to be 90 degrees
  • 00:33:34
    and sine 90 is one
  • 00:33:36
    so f is simply equal to b q v
  • 00:33:43
    b
  • 00:33:44
    is the magnetic field in tesla that's
  • 00:33:46
    two times ten to the minus four teslas
  • 00:33:50
    q is the charge of just one proton
  • 00:33:53
    the charge of a proton
  • 00:33:55
    is 1.6 times 10 to the negative 19
  • 00:33:57
    coulombs
  • 00:33:58
    and that's something you just got to
  • 00:34:00
    know
  • 00:34:04
    and the speed of the proton is 4 times
  • 00:34:07
    10 to the 6 meters per second
  • 00:34:10
    so we just gotta
  • 00:34:12
    multiply these three numbers
  • 00:34:24
    and so you should get
  • 00:34:26
    1.28
  • 00:34:28
    times 10 to the minus 16
  • 00:34:31
    newtons
  • 00:34:33
    so that's the magnetic force acting on
  • 00:34:35
    the proton
  • 00:34:39
    now let's talk about a proton
  • 00:34:42
    so let's say if we have a proton
  • 00:34:45
    and it's moving towards the right
  • 00:34:49
    and the magnetic field
  • 00:34:51
    is directed everywhere
  • 00:34:53
    into the page
  • 00:34:56
    so let's say there's an x
  • 00:34:58
    everywhere
  • 00:35:04
    what's going to happen
  • 00:35:05
    where is the magnetic force
  • 00:35:08
    so if you direct your thumb towards the
  • 00:35:10
    right and your four fingers into the
  • 00:35:12
    page
  • 00:35:13
    the magnetic force will be directed
  • 00:35:17
    north
  • 00:35:19
    whenever force and velocity are
  • 00:35:21
    perpendicular to each other
  • 00:35:23
    the object will turn
  • 00:35:26
    so it's going to go this way
  • 00:35:28
    if force and velocity are in the same
  • 00:35:30
    direction the object will speed up
  • 00:35:32
    if force and velocity are in the
  • 00:35:34
    opposite direction the object slows down
  • 00:35:36
    but if they're perpendicular the object
  • 00:35:39
    will turn
  • 00:35:42
    and eventually the particle
  • 00:35:45
    is going to be moving in a direction of
  • 00:35:47
    the force
  • 00:35:48
    but
  • 00:35:49
    now that the velocity is directed north
  • 00:35:52
    the force is going to change
  • 00:35:54
    if you use the right hand rule again if
  • 00:35:56
    you direct your four fingers into the
  • 00:35:57
    page
  • 00:35:58
    the velocity north
  • 00:36:00
    the force will be directed
  • 00:36:02
    west
  • 00:36:03
    and so what's going to happen is
  • 00:36:05
    this particle the proton is going to
  • 00:36:08
    move
  • 00:36:09
    in a circle
  • 00:36:11
    i haven't drawn a knife circle because i
  • 00:36:12
    ran out of space but you get the picture
  • 00:36:15
    when it's at the top
  • 00:36:16
    the force is going to be directed
  • 00:36:19
    south
  • 00:36:20
    as it moves
  • 00:36:22
    to the left
  • 00:36:27
    and as the proton
  • 00:36:28
    moves south
  • 00:36:30
    the force will be directed
  • 00:36:32
    east
  • 00:36:33
    so notice that
  • 00:36:35
    for a moving charge particle
  • 00:36:37
    the magnetic force behaves as a
  • 00:36:39
    centripetal force or a sensor seeking
  • 00:36:42
    force
  • 00:36:43
    now what if we have an electron
  • 00:36:48
    if we had an electron
  • 00:36:50
    the situation will be opposite
  • 00:36:55
    as the electron is moving in the same
  • 00:36:57
    direction
  • 00:36:58
    as the proton
  • 00:37:00
    it's going to fill a force in the
  • 00:37:02
    opposite direction
  • 00:37:03
    so the proton felt a force
  • 00:37:06
    that was directed north the electron
  • 00:37:08
    will feel a force directed in the
  • 00:37:10
    opposite direction at itself
  • 00:37:13
    so as the proton
  • 00:37:15
    moves in the
  • 00:37:18
    counterclockwise direction
  • 00:37:20
    the electron will move in a clockwise
  • 00:37:23
    direction
  • 00:37:25
    so they will move in an opposite
  • 00:37:27
    direction
  • 00:37:29
    now how can we calculate the radius of
  • 00:37:31
    curvature
  • 00:37:33
    that a proton or electron
  • 00:37:35
    might travel in a circle how can we
  • 00:37:37
    figure out the radius
  • 00:37:39
    if you ever get a question like this
  • 00:37:41
    what you need to do is set the
  • 00:37:42
    centripetal force equal to
  • 00:37:45
    the magnetic force
  • 00:37:47
    let's call fb the magnetic force f
  • 00:37:49
    status and triple force
  • 00:37:51
    the centripetal force
  • 00:37:53
    is
  • 00:37:55
    provided by the magnetic force
  • 00:37:58
    in this particular example the
  • 00:38:00
    centripetal force is equal to mv squared
  • 00:38:02
    divided by r which is the radius of the
  • 00:38:04
    circle the magnetic force is equal to
  • 00:38:06
    bqv
  • 00:38:08
    now
  • 00:38:10
    because the magnetic field and the
  • 00:38:11
    velocity are perpendicular we don't have
  • 00:38:14
    to worry about the sine portion of this
  • 00:38:15
    equation
  • 00:38:17
    now we're going to do is multiply both
  • 00:38:18
    sides
  • 00:38:19
    by 1 over v
  • 00:38:22
    so on the right side
  • 00:38:23
    v will cancel
  • 00:38:25
    on the left side
  • 00:38:26
    because we have a v squared one of them
  • 00:38:28
    will remain
  • 00:38:30
    so mv divided by r is equal to bq
  • 00:38:35
    so multiplying both sides by r
  • 00:38:39
    we have this equation
  • 00:38:42
    so make sure you uh write down this
  • 00:38:44
    equation mv is equal to bqr
  • 00:38:48
    because in this format you can solve for
  • 00:38:50
    anything
  • 00:38:52
    so let's say if we want to find the
  • 00:38:53
    radius of the circle
  • 00:38:55
    all we need to do is divide both sides
  • 00:38:57
    by
  • 00:38:58
    bq
  • 00:39:00
    so the radius of the circle
  • 00:39:02
    is equal to the mass of the charged
  • 00:39:04
    particle
  • 00:39:05
    times the velocity
  • 00:39:06
    divided by the magnetic field
  • 00:39:08
    times the charge
  • 00:39:14
    number six
  • 00:39:15
    a proton
  • 00:39:16
    moves with a speed of five times ten to
  • 00:39:19
    six meters per second
  • 00:39:20
    in a plane perpendicular to a magnetic
  • 00:39:22
    field of 2.5 tesla
  • 00:39:26
    calculate the radius
  • 00:39:27
    of its circular path
  • 00:39:30
    so here's the proton
  • 00:39:34
    and if it's moving perpendicular to a
  • 00:39:36
    magnetic field it's going to move in a
  • 00:39:37
    circle
  • 00:39:39
    our goal
  • 00:39:40
    is to calculate the radius of that
  • 00:39:42
    circle
  • 00:39:45
    so r
  • 00:39:47
    is equal to
  • 00:39:49
    based on the equation that we had before
  • 00:39:50
    it's mv
  • 00:39:52
    divided by bq
  • 00:39:54
    so what is the mass of a proton
  • 00:39:57
    now the problem doesn't give it to you
  • 00:39:59
    which means you can either look it up
  • 00:40:00
    online
  • 00:40:01
    or you can look it up in the reference
  • 00:40:03
    section of your textbook
  • 00:40:05
    the mass of a proton is about 1.673
  • 00:40:10
    times 10 to the minus 27 kilograms
  • 00:40:14
    the speed
  • 00:40:15
    which is given that's 5 times 10 to 6
  • 00:40:18
    meters per second
  • 00:40:20
    the strength of the magnetic field is
  • 00:40:22
    2.5 tesla
  • 00:40:24
    and the charge of a proton
  • 00:40:27
    which is
  • 00:40:28
    the same as that of an electron but the
  • 00:40:30
    opposite sign it's 1.6
  • 00:40:34
    times 10 to the negative 19 coulombs
  • 00:40:38
    so if we type these numbers into the
  • 00:40:39
    calculator
  • 00:40:43
    we should get
  • 00:40:52
    point zero two
  • 00:40:54
    zero nine meters
  • 00:40:56
    which
  • 00:40:58
    is equal to
  • 00:41:01
    two point zero nine centimeters
  • 00:41:04
    so that's the radius of the path that
  • 00:41:07
    the protons gonna travel in
  • 00:41:09
    and so that's how you could find it
  • 00:41:12
    now what about part b
  • 00:41:13
    what is the energy
  • 00:41:15
    of the proton and electron volts
  • 00:41:18
    how can we find the answer to that
  • 00:41:19
    question
  • 00:41:22
    well first
  • 00:41:23
    we need to find the energy in joules
  • 00:41:27
    a moving object has kinetic energy any
  • 00:41:30
    object in motion
  • 00:41:31
    contains kinetic energy so we've got to
  • 00:41:33
    find the kinetic energy of the proton
  • 00:41:36
    which is one-half mv squared
  • 00:41:39
    so we know the mass of a proton
  • 00:41:42
    it's 1.673 times 10 to the negative 27
  • 00:41:47
    and we also have the speed
  • 00:41:49
    5 times 10 to the 6 meters per second
  • 00:41:51
    squared
  • 00:42:00
    so this is equal to
  • 00:42:03
    2.09 times 10 to the negative 14
  • 00:42:08
    joules
  • 00:42:09
    now once you have the energy in joules
  • 00:42:12
    you can convert it to electron volts
  • 00:42:14
    electron volts is basically another unit
  • 00:42:17
    of energy
  • 00:42:18
    it's very useful for
  • 00:42:20
    small particles like protons and
  • 00:42:22
    electrons
  • 00:42:25
    one electron volt
  • 00:42:30
    is equal to 1.6 times 10
  • 00:42:34
    to the negative 19 joules
  • 00:42:36
    we'll talk about why later
  • 00:42:37
    but that's what it's equal to
  • 00:42:40
    and so you just got to convert it to
  • 00:42:41
    electron volts
  • 00:42:43
    and this will give you the answer
  • 00:42:49
    which is a 130 000
  • 00:42:52
    and 703 electron volts
  • 00:42:57
    so that's it for this problem
  • 00:43:02
    so why is it that
  • 00:43:04
    one electron volt
  • 00:43:07
    is equal to
  • 00:43:08
    1.6
  • 00:43:11
    times 10 to the negative 19 joules
  • 00:43:14
    why is that the case
  • 00:43:18
    electric potential which is measured in
  • 00:43:21
    volts
  • 00:43:22
    is equal to
  • 00:43:24
    the electric potential energy
  • 00:43:27
    which is measured in joules
  • 00:43:29
    divided by the charge
  • 00:43:34
    so the unit volt one volt
  • 00:43:36
    is equal to one joule
  • 00:43:39
    per one column
  • 00:43:44
    so therefore
  • 00:43:46
    an electron has a charge of 1.6
  • 00:43:50
    times 10 to the negative 19 coulombs
  • 00:43:53
    it's negative but let's ignore the
  • 00:43:54
    negative sign
  • 00:43:55
    that's the charge of an electron
  • 00:43:59
    and
  • 00:44:00
    an electron that has one volt
  • 00:44:05
    or one ev that's one electron volt an
  • 00:44:07
    electron with one volt
  • 00:44:09
    will have an energy of 1.6 times 10 to
  • 00:44:12
    the negative 19
  • 00:44:13
    joules
  • 00:44:16
    volt is basically the ratio between
  • 00:44:18
    joules and coulombs
  • 00:44:20
    so if you have a charged particle that
  • 00:44:22
    has one joule and one clue
  • 00:44:25
    its voltage is one volt
  • 00:44:28
    or the electric potential is one volt
  • 00:44:30
    voltage is really work per unit charge
  • 00:44:33
    electric potential is energy per unit
  • 00:44:34
    charge
  • 00:44:37
    now if we have a charged particle that
  • 00:44:40
    has an energy of 1.6 times 10 to
  • 00:44:42
    negative 19 joules
  • 00:44:43
    and a charge
  • 00:44:45
    of that many coulombs
  • 00:44:47
    then these two
  • 00:44:48
    will cancel and the electric potential
  • 00:44:51
    will be 1 volt
  • 00:44:53
    which is the case of an electron
  • 00:44:55
    an electron
  • 00:44:57
    has a charge of 1.6 times 10 to negative
  • 00:45:00
    19 coulombs
  • 00:45:02
    and if that electron has an energy of
  • 00:45:04
    one electron volt its voltage is one
  • 00:45:06
    volt which means its energy
  • 00:45:09
    is equal to this number
  • 00:45:11
    and that's why one electron volt is 1.6
  • 00:45:14
    times 10 to the negative 19 joules
  • 00:45:16
    just in case you're wondering
  • 00:45:21
    now let's say if we have two wires
  • 00:45:25
    parallel to each other
  • 00:45:28
    and let's say that there's a current
  • 00:45:31
    in wire one
  • 00:45:32
    and in wire two
  • 00:45:34
    and these two currents are in the same
  • 00:45:35
    direction
  • 00:45:38
    will these two wires attract each other
  • 00:45:40
    or will they repel
  • 00:45:42
    it turns out that
  • 00:45:44
    these two wires will attract each other
  • 00:45:47
    if they have a current in the same
  • 00:45:48
    direction
  • 00:45:50
    now if there's two wires
  • 00:45:58
    with the opposite current
  • 00:46:01
    then the situation is different
  • 00:46:05
    they will not feel a force of attraction
  • 00:46:08
    rather
  • 00:46:09
    they will repel each other
  • 00:46:14
    so why does that happen
  • 00:46:17
    why do we have a force of attraction in
  • 00:46:19
    the first case
  • 00:46:22
    but in the second scenario
  • 00:46:24
    these two repel
  • 00:46:30
    what's going on here
  • 00:46:34
    wire one
  • 00:46:36
    creates a magnetic field because it has
  • 00:46:38
    a moving charge it has a current
  • 00:46:41
    and that magnetic field exerts a force
  • 00:46:44
    on wire two
  • 00:46:47
    now let's focus on wire one
  • 00:46:49
    the current is going
  • 00:46:51
    north
  • 00:46:52
    if you use the first right hand rule
  • 00:46:54
    that we talked about earlier in the
  • 00:46:55
    video where you curl your hand around
  • 00:46:57
    the wire and your thumb is in the
  • 00:46:59
    direction of the current
  • 00:47:03
    the magnetic force created by wire one
  • 00:47:06
    will be
  • 00:47:07
    out of the page
  • 00:47:08
    on the left side
  • 00:47:11
    but it's going to be into the page
  • 00:47:13
    on the right side
  • 00:47:16
    so wire 2 is on the right side of wire 1
  • 00:47:19
    therefore wire 2
  • 00:47:21
    sees a magnetic field that's going into
  • 00:47:23
    the page now using the second right hand
  • 00:47:25
    rule that we talked about
  • 00:47:27
    what you want to do
  • 00:47:29
    is you want your fingers
  • 00:47:31
    to point
  • 00:47:32
    into the page
  • 00:47:38
    but you want your thumb
  • 00:47:41
    pointing north in the direction of the
  • 00:47:43
    current
  • 00:47:48
    so here's your thumb
  • 00:47:50
    it follows i2
  • 00:47:52
    the magnetic field
  • 00:47:55
    which is b1 that's created by i1
  • 00:47:57
    you want that to be
  • 00:47:59
    in the page
  • 00:48:03
    and what's going to happen is
  • 00:48:05
    the force is going to come out of the
  • 00:48:06
    palm of your hand
  • 00:48:08
    and it turns out
  • 00:48:10
    that force is directed towards
  • 00:48:12
    wire one
  • 00:48:14
    so it's a force of attraction
  • 00:48:17
    and so anytime you have two wires
  • 00:48:20
    with the current going in the same
  • 00:48:22
    direction
  • 00:48:23
    it's going to create an attractive force
  • 00:48:25
    the two wires will be tracked to each
  • 00:48:27
    other
  • 00:48:28
    if the current
  • 00:48:29
    is in the opposite direction
  • 00:48:32
    then the two wires will repel each other
  • 00:48:36
    now how can we calculate
  • 00:48:38
    the magnitude of the force between the
  • 00:48:40
    two wires
  • 00:48:43
    so we said that the first wire wire one
  • 00:48:46
    creates a magnetic field
  • 00:48:48
    that causes the second wire
  • 00:48:51
    to be attracted to the first if they're
  • 00:48:52
    moving in the same direction
  • 00:48:54
    and
  • 00:48:55
    also the second wire creates a magnetic
  • 00:48:57
    field
  • 00:48:58
    that exerts a force on the first wire
  • 00:49:01
    causing the first wire to move towards
  • 00:49:03
    the second if the currents are in the
  • 00:49:04
    same direction
  • 00:49:07
    so
  • 00:49:08
    let's start with this equation wire one
  • 00:49:10
    creates a magnetic field b1 which is u0
  • 00:49:14
    times i 1
  • 00:49:16
    over 2 pi r
  • 00:49:17
    where r
  • 00:49:19
    is the distance
  • 00:49:20
    between the two wires
  • 00:49:25
    so wire 1
  • 00:49:27
    which produces a current one
  • 00:49:29
    will generate a magnetic field
  • 00:49:33
    that is at wire two
  • 00:49:36
    and so r is the distance between the two
  • 00:49:37
    wires
  • 00:49:40
    now we need to use the other equation
  • 00:49:42
    that is the force that
  • 00:49:46
    acts on a current carrying wire that's f
  • 00:49:48
    is equal to
  • 00:49:50
    ilb sine theta but we're not going to be
  • 00:49:53
    worried about the angle in this problem
  • 00:49:56
    so the force on wire two which we'll
  • 00:49:58
    call f2
  • 00:49:59
    is equal to the current
  • 00:50:01
    on that wire times the length
  • 00:50:03
    times the magnetic field created by wire
  • 00:50:06
    one
  • 00:50:07
    so what we're going to do now is replace
  • 00:50:09
    b1
  • 00:50:10
    with mu 0 i 1 over 2 pi r
  • 00:50:17
    so it's going to be i 2
  • 00:50:19
    times l
  • 00:50:20
    times mu 0 i 1
  • 00:50:23
    over 2 pi r
  • 00:50:25
    so that's how you can calculate
  • 00:50:27
    the force between the two wires
  • 00:50:29
    f2 and f1 they have the same magnitude
  • 00:50:35
    number seven
  • 00:50:36
    what is the magnitude and direction of
  • 00:50:38
    the force between two parallel wires
  • 00:50:41
    that are 30 meters long
  • 00:50:43
    and two centimeters apart
  • 00:50:46
    each carrying a current of 50 amps in
  • 00:50:48
    the same direction
  • 00:50:50
    let's draw a picture
  • 00:50:52
    so we have two wires
  • 00:50:58
    and they have the same current
  • 00:51:03
    so i1
  • 00:51:04
    and i2
  • 00:51:06
    both equal 50 amps
  • 00:51:09
    now we have the length of the wire which
  • 00:51:11
    is l
  • 00:51:14
    and the left of both wires
  • 00:51:17
    is 30 meters
  • 00:51:19
    and the distance between them which is r
  • 00:51:22
    that's two centimeters
  • 00:51:24
    which is equivalent to point zero two
  • 00:51:27
    meters
  • 00:51:30
    now for these type of problems there's
  • 00:51:32
    only two answers for the direction
  • 00:51:35
    either the force is attractive
  • 00:51:38
    or
  • 00:51:39
    they will repel
  • 00:51:40
    because the currents are in the same
  • 00:51:42
    direction we have a force of attraction
  • 00:51:45
    and so that answers the direction of the
  • 00:51:47
    force
  • 00:51:48
    it's simply attraction
  • 00:51:50
    now all we need to do is find the
  • 00:51:52
    magnitude
  • 00:51:53
    so let's use the formula
  • 00:51:55
    f
  • 00:51:56
    is equal to mu zero
  • 00:51:59
    times i1 times i2
  • 00:52:03
    times l which is the length of the wires
  • 00:52:06
    divided by 2 pi
  • 00:52:08
    times r
  • 00:52:10
    so it's 4 pi times 10 to the minus 7
  • 00:52:14
    times 50 times 50 which we can write as
  • 00:52:16
    50 squared
  • 00:52:18
    times l
  • 00:52:19
    which is 30 meters
  • 00:52:21
    divided by 2 pi
  • 00:52:24
    times the radius of 0.02 meters
  • 00:52:27
    now 4 pi divided by 2 pi
  • 00:52:30
    is going to be 2.
  • 00:52:32
    so it's 2 times 10 to the minus 7
  • 00:52:35
    times 50 squared
  • 00:52:38
    times 30
  • 00:52:39
    divided by .02
  • 00:52:44
    so the force between these two wires
  • 00:52:47
    is 0.75 newtons
  • 00:52:53
    and a direction
  • 00:52:55
    is a force of attraction
  • 00:52:57
    these two forces
  • 00:52:59
    will be pointed towards each other
  • 00:53:02
    now the next thing that we need to talk
  • 00:53:04
    about
  • 00:53:05
    is ampere's law
  • 00:53:09
    ampere's law describes the relationship
  • 00:53:11
    between
  • 00:53:13
    the current and magnetic field produced
  • 00:53:15
    by that current
  • 00:53:17
    perhaps you've seen this equation
  • 00:53:20
    the sum of all the magnetic fields that
  • 00:53:22
    is parallel
  • 00:53:24
    to any uh segments that the magnetic
  • 00:53:27
    field passes through
  • 00:53:29
    that's going to be equal to
  • 00:53:32
    mu zero
  • 00:53:33
    times the current enclosed
  • 00:53:35
    by the path that the magnetic field
  • 00:53:37
    makes and has to be a closed path
  • 00:53:41
    a good example
  • 00:53:42
    is using
  • 00:53:44
    a wire
  • 00:53:45
    so let's say if we have a current that
  • 00:53:47
    passes through this wire
  • 00:53:52
    now this current will create a magnetic
  • 00:53:54
    field
  • 00:53:54
    that travels
  • 00:53:56
    around the wire in a circular path
  • 00:54:01
    so if we take
  • 00:54:03
    the magnetic field
  • 00:54:05
    and multiply it by the path that is
  • 00:54:08
    parallel to it let's call it delta l
  • 00:54:14
    we can use that to
  • 00:54:16
    get an equation that will give us the
  • 00:54:18
    magnetic field created by this wire
  • 00:54:21
    so the path that the magnetic field
  • 00:54:23
    travels is basically the path of a
  • 00:54:25
    circle
  • 00:54:27
    so delta l
  • 00:54:28
    is really
  • 00:54:30
    two pi r
  • 00:54:31
    if we add up all the small segments so
  • 00:54:33
    if we add up this segment plus that
  • 00:54:35
    segment plus that segment we're going to
  • 00:54:38
    get the circumference of a circle which
  • 00:54:39
    is 2 pi r
  • 00:54:42
    so it's going to be b
  • 00:54:43
    times 2 pi r which equals mu is not over
  • 00:54:46
    i i mean times i
  • 00:54:49
    so the magnetic field
  • 00:54:50
    if you divide both sides by two pi r
  • 00:54:53
    we get this familiar equation mu zero
  • 00:54:55
    times i
  • 00:54:56
    over two pi r
  • 00:55:00
    now it's important to understand that
  • 00:55:02
    the current in this equation
  • 00:55:04
    is the current that is enclosed
  • 00:55:07
    by this loop
  • 00:55:09
    so that loop has an area
  • 00:55:11
    and
  • 00:55:13
    the current that passes through that
  • 00:55:15
    area that's this current is the current
  • 00:55:17
    enclosed by that circular loop
  • 00:55:20
    and so that's how you can use ampere's
  • 00:55:22
    law to get the equation
  • 00:55:24
    of a magnetic field by a current
  • 00:55:25
    carrying wire
  • 00:55:27
    now let's use ampere's law to come up
  • 00:55:29
    with an equation
  • 00:55:31
    for a solenoid
  • 00:55:32
    a solenoid is basically
  • 00:55:35
    a device with many loops of wire
  • 00:55:39
    and the reason why
  • 00:55:41
    it's advantageous to create a solenoid
  • 00:55:45
    anytime you create a loop of wire
  • 00:55:47
    whenever you have a current the magnetic
  • 00:55:49
    field that is at the center of the wire
  • 00:55:51
    is very strong
  • 00:55:53
    and for every loop that you add you
  • 00:55:55
    increase the strength of the magnetic
  • 00:55:56
    field
  • 00:55:58
    inside the wire
  • 00:56:00
    and so solenoids are very useful for
  • 00:56:02
    creating powerful magnetic fields
  • 00:56:07
    outside of the loop the magnetic field
  • 00:56:10
    is weak
  • 00:56:12
    so the magnetic field will travel
  • 00:56:14
    in a circular pattern
  • 00:56:17
    so to calculate the
  • 00:56:19
    or to derive the formula for a solenoid
  • 00:56:21
    we need to create a path
  • 00:56:24
    of the magnetic field
  • 00:56:26
    so let's draw a rectangular path
  • 00:56:36
    so let's say this is a
  • 00:56:39
    b
  • 00:56:40
    c
  • 00:56:41
    and d
  • 00:56:45
    so what we need to do
  • 00:56:47
    is add up the magnetic field that is
  • 00:56:50
    parallel to each segment
  • 00:56:54
    so that's going to be the magnetic field
  • 00:56:56
    times the left of segment
  • 00:56:59
    ac
  • 00:57:01
    plus the magnetic field
  • 00:57:03
    times the length of segment
  • 00:57:05
    cd
  • 00:57:07
    plus the magnetic field times the length
  • 00:57:09
    of segment
  • 00:57:11
    db
  • 00:57:12
    plus the magnetic field times the length
  • 00:57:14
    of segment
  • 00:57:16
    ba
  • 00:57:20
    now
  • 00:57:21
    let's focus on segment bd
  • 00:57:24
    segment bd is outside of the solenoid
  • 00:57:27
    and the magnetic field is very weak
  • 00:57:29
    outside of the solenoid so
  • 00:57:31
    we can say that
  • 00:57:32
    the contribution for
  • 00:57:35
    uh bl
  • 00:57:37
    and segment bd is very small
  • 00:57:40
    so it's negligible
  • 00:57:44
    now ba
  • 00:57:45
    and dc
  • 00:57:47
    they're perpendicular
  • 00:57:49
    to the magnetic field that is inside the
  • 00:57:51
    solenoid
  • 00:57:53
    that magnetic field is much stronger
  • 00:57:55
    than the magnetic field on the outside
  • 00:57:59
    so it's because it's perpendicular to
  • 00:58:00
    the magnetic field that's inside the
  • 00:58:02
    solenoid
  • 00:58:04
    its contribution is going to be
  • 00:58:05
    negligible
  • 00:58:06
    so we can eliminate ba
  • 00:58:09
    and dc
  • 00:58:16
    so therefore
  • 00:58:17
    all we have is
  • 00:58:19
    the segment that is parallel
  • 00:58:22
    to the magnetic field
  • 00:58:24
    that is inside the solenoid that's ac
  • 00:58:28
    segment ac
  • 00:58:30
    is the most important segment because
  • 00:58:32
    one it's parallel to the magnetic field
  • 00:58:34
    that is inside the solenoid
  • 00:58:37
    and that magnetic field is the strongest
  • 00:58:38
    one compared to the ones that are
  • 00:58:40
    outside of it
  • 00:58:41
    and so
  • 00:58:43
    bl
  • 00:58:46
    in segment ac
  • 00:58:47
    will have the greatest contribution
  • 00:58:49
    towards the sum
  • 00:58:51
    of all the
  • 00:58:52
    magnetic fields that is parallel to each
  • 00:58:54
    segment
  • 00:58:58
    so now using ampere's law
  • 00:59:02
    which is basically
  • 00:59:03
    this equation
  • 00:59:05
    we can now replace this term with b
  • 00:59:07
    times l
  • 00:59:09
    where l
  • 00:59:10
    is simply the length of the solenoid
  • 00:59:13
    the length of the solenoid being
  • 00:59:14
    segment ac but we'll just call it l
  • 00:59:19
    so bl is equal to mu zero times i
  • 00:59:22
    but this is four
  • 00:59:24
    if we only have one loop
  • 00:59:26
    let's say if you have a wire with a
  • 00:59:28
    current of
  • 00:59:30
    ten if you have one loop then the
  • 00:59:33
    enclosed current is 10 amps but it turns
  • 00:59:35
    out that if you add another loop
  • 00:59:38
    the
  • 00:59:40
    current enclosed by the magnetic field
  • 00:59:45
    it's going to be twice as much even
  • 00:59:47
    though 10 amps is flowing through the
  • 00:59:48
    wire
  • 00:59:49
    the enclosed current is now 20.
  • 00:59:51
    and if you add another loop
  • 00:59:53
    the enclosed current is 30 even though
  • 00:59:55
    10 amps is still flowing through the
  • 00:59:57
    same wire
  • 00:59:58
    so therefore we need to add n to this
  • 01:00:01
    equation
  • 01:00:03
    because the enclosed current increases
  • 01:00:06
    so the enclosed current is basically
  • 01:00:08
    the current that flows in one loop times
  • 01:00:10
    the number of loops
  • 01:00:15
    so now all we need to do is divide both
  • 01:00:17
    sides by l
  • 01:00:20
    so b is equal to mu zero
  • 01:00:23
    times n
  • 01:00:24
    times i divided by l
  • 01:00:27
    lower case n is equal to capital n
  • 01:00:30
    over l
  • 01:00:31
    capital n represents the number of terms
  • 01:00:34
    or loops
  • 01:00:36
    l
  • 01:00:37
    is the length in meters
  • 01:00:39
    so lower case n is the number of loops
  • 01:00:43
    or turns
  • 01:00:45
    per meter
  • 01:00:49
    so we're going to replace capital n over
  • 01:00:51
    l with lowercase n
  • 01:00:53
    so now we have the equation of a
  • 01:00:54
    solenoid
  • 01:00:57
    b is equal to mu zero
  • 01:00:59
    times n times i
  • 01:01:02
    so the magnetic field
  • 01:01:04
    produced by a solenoid
  • 01:01:07
    is proportional to the current that
  • 01:01:08
    passes through
  • 01:01:10
    as the current increases
  • 01:01:12
    the strength of the magnetic field will
  • 01:01:14
    increase
  • 01:01:15
    the second way to increase the magnetic
  • 01:01:18
    field
  • 01:01:19
    is to increase the number of turns
  • 01:01:22
    if you can increase the amount of turns
  • 01:01:24
    per unit left or per meter
  • 01:01:27
    the strength for the magnetic field will
  • 01:01:28
    greatly increase
  • 01:01:32
    so you want to increase the number of
  • 01:01:33
    turns you want to increase the current
  • 01:01:35
    but you also want to decrease the length
  • 01:01:37
    if you can decrease the length
  • 01:01:40
    the magnetic field will increase as well
  • 01:01:43
    so the magnetic field is directly
  • 01:01:44
    related to the current
  • 01:01:47
    it's related to the number of turns and
  • 01:01:49
    it's inversely related to length
  • 01:01:53
    number eight
  • 01:01:54
    a solenoid has a length of 15
  • 01:01:57
    centimeters
  • 01:01:58
    and a total of 800 turns of wire
  • 01:02:01
    calculate the strength of the magnetic
  • 01:02:03
    field at its center if the solenoid
  • 01:02:05
    carries the current of 5 amps
  • 01:02:07
    so first
  • 01:02:09
    let's calculate lower case n which is
  • 01:02:11
    capital n over l
  • 01:02:14
    so there's 800 turns
  • 01:02:17
    and the length of the wire is 15
  • 01:02:19
    centimeters but we need to convert that
  • 01:02:21
    to meters
  • 01:02:22
    so let's divide by 100.
  • 01:02:24
    15 divided by 100 is 0.15
  • 01:02:27
    or 0.15 meters
  • 01:02:31
    so 800
  • 01:02:33
    divided by point fifteen
  • 01:02:36
    that's equal to
  • 01:02:38
    five thousand
  • 01:02:39
    three hundred and thirty three
  • 01:02:42
    turns per meter
  • 01:02:44
    or loops per meter
  • 01:02:47
    so now to calculate the strength of the
  • 01:02:48
    magnetic field
  • 01:02:50
    it's mu zero times n times i
  • 01:02:54
    so it's four pi times ten to minus seven
  • 01:02:58
    times the number of turns per meter
  • 01:03:00
    which is five 5333
  • 01:03:03
    times the current which is 5 amps
  • 01:03:16
    so the strength for the magnetic field
  • 01:03:18
    at the center it's about .0335
  • 01:03:22
    tesla
  • 01:03:25
    and so
  • 01:03:26
    that's it for this problem
  • 01:03:31
    now what's going to happen if we have
  • 01:03:34
    a current carrying loop
  • 01:03:36
    inside
  • 01:03:38
    a magnetic field
  • 01:03:46
    in a magnetic field
  • 01:03:48
    the loop is going to rotate it's going
  • 01:03:50
    to produce a torque
  • 01:03:52
    so let's say on the left side there's a
  • 01:03:53
    current
  • 01:03:55
    and on the right side there's a current
  • 01:03:59
    so the current travels clockwise
  • 01:04:02
    in this loop and is also a magnetic
  • 01:04:04
    field
  • 01:04:05
    that is directed
  • 01:04:06
    east
  • 01:04:10
    now what type of force will we have on
  • 01:04:12
    the left side of the wire or the metal
  • 01:04:14
    loop
  • 01:04:16
    so using the right hand rule
  • 01:04:19
    place your thumb
  • 01:04:20
    going up
  • 01:04:21
    and your four fingers in the direction
  • 01:04:23
    of the magnetic field
  • 01:04:30
    and if you do that notice that the palm
  • 01:04:33
    of your hand
  • 01:04:34
    opens into the page
  • 01:04:37
    and so that's where the force is going
  • 01:04:38
    to be
  • 01:04:42
    so the magnetic field will exert a force
  • 01:04:44
    on the left side of the loop
  • 01:04:47
    going into the page therefore on the
  • 01:04:49
    right side of the loop the current is
  • 01:04:50
    reversed
  • 01:04:51
    so it must be out of the page
  • 01:04:56
    so therefore this
  • 01:04:58
    loop is going to turn
  • 01:05:00
    it's going to turn this way
  • 01:05:02
    and then into the page
  • 01:05:03
    on the other side
  • 01:05:08
    now let's talk about how to
  • 01:05:10
    derive an equation
  • 01:05:12
    for the torque of this current carrying
  • 01:05:14
    loop
  • 01:05:17
    f is equal to
  • 01:05:18
    ilb
  • 01:05:20
    it's ilb sine theta but the force is
  • 01:05:23
    perpendicular both to the current and
  • 01:05:25
    magnetic field
  • 01:05:26
    so sine 90 is one
  • 01:05:29
    so the force acting on the right side
  • 01:05:31
    and on the left side of the loop
  • 01:05:33
    it's going to be
  • 01:05:34
    ilb there's no force on the top section
  • 01:05:37
    in the bottom section of the loop
  • 01:05:40
    the reason for that is because the
  • 01:05:41
    current
  • 01:05:43
    is parallel to the magnetic field
  • 01:05:45
    and
  • 01:05:46
    whenever the current is parallel to the
  • 01:05:47
    magnetic field
  • 01:05:49
    there's going to be no magnetic force it
  • 01:05:50
    has to be perpendicular to it
  • 01:05:52
    so there's only a force on the left side
  • 01:05:54
    and on the right side
  • 01:05:55
    not on the top or bottom section of the
  • 01:05:57
    loop
  • 01:06:00
    so since we're focused on
  • 01:06:02
    this side that's where the magnetic
  • 01:06:04
    force will be exerted on
  • 01:06:07
    l
  • 01:06:08
    represents the length of that section
  • 01:06:13
    so that's this l b is the magnetic field
  • 01:06:17
    the width of the loop
  • 01:06:18
    let's call it
  • 01:06:20
    w
  • 01:06:25
    let's put that here
  • 01:06:30
    now we need to calculate the torque
  • 01:06:34
    this force will create a torque and this
  • 01:06:36
    force will create another torque that's
  • 01:06:38
    additive
  • 01:06:40
    so the net torque
  • 01:06:42
    is going to be t1
  • 01:06:44
    plus t2
  • 01:06:46
    and just to refresh you on how to
  • 01:06:48
    calculate torque
  • 01:06:50
    torque is the product of the force
  • 01:06:53
    times the level arm
  • 01:06:55
    so it's f times r
  • 01:06:56
    and if the force is at an angle
  • 01:06:59
    is going to be fr
  • 01:07:01
    sine theta
  • 01:07:06
    so
  • 01:07:07
    t1
  • 01:07:09
    and t2
  • 01:07:11
    is going to be
  • 01:07:13
    f1 times
  • 01:07:14
    r1 times sine theta
  • 01:07:17
    plus
  • 01:07:19
    f2 times r2
  • 01:07:21
    sine theta
  • 01:07:23
    so what exactly is r
  • 01:07:25
    in this example
  • 01:07:27
    we know that r
  • 01:07:28
    is the distance between
  • 01:07:30
    where the force is applied
  • 01:07:32
    and the axis of rotation that's where
  • 01:07:35
    the object moves around
  • 01:07:38
    so in this particular problem
  • 01:07:41
    the axis of rotation
  • 01:07:44
    is here
  • 01:07:47
    so which means that r
  • 01:07:50
    is half
  • 01:07:52
    of w
  • 01:07:54
    so r
  • 01:07:56
    is w divided by two
  • 01:07:59
    so now
  • 01:08:00
    we can replace f with ilb
  • 01:08:04
    and we can replace
  • 01:08:06
    r
  • 01:08:07
    with w over two
  • 01:08:11
    and then t2 is going to be the same
  • 01:08:13
    thing it's the same current
  • 01:08:15
    times
  • 01:08:16
    the same length they both have the same
  • 01:08:18
    life
  • 01:08:19
    they're both exposed to the same
  • 01:08:21
    magnetic field
  • 01:08:22
    and r is the same r1 and r2 is the same
  • 01:08:28
    so we no longer need this picture
  • 01:08:31
    so what i'm going to do is i'm going to
  • 01:08:33
    factor out ilb and sine theta
  • 01:08:36
    so t
  • 01:08:37
    is ilb
  • 01:08:40
    sine theta
  • 01:08:42
    w over two plus w over two half plus
  • 01:08:46
    half is a whole
  • 01:08:48
    so t is equal to i
  • 01:08:50
    l w
  • 01:08:51
    b sine theta
  • 01:08:54
    now for the rectangular loop that we
  • 01:08:55
    have
  • 01:08:57
    which has
  • 01:08:59
    a length l
  • 01:09:01
    and a with w
  • 01:09:05
    the area of the loop is basically the
  • 01:09:07
    length times the width
  • 01:09:09
    so a
  • 01:09:10
    is l times w so let's replace lw with a
  • 01:09:15
    so the torque of a single loop
  • 01:09:17
    is the current
  • 01:09:18
    multiplied by the area
  • 01:09:20
    times the strength of the magnetic field
  • 01:09:23
    times sine of theta
  • 01:09:25
    now what if we have many loops
  • 01:09:27
    if you have two loops the torque is
  • 01:09:30
    going to be twice as strong
  • 01:09:31
    three loops three times strong so we
  • 01:09:34
    need to add n to this equation for the
  • 01:09:36
    number of loops so it's n
  • 01:09:38
    i a b
  • 01:09:40
    sine theta
  • 01:09:42
    by the way the quantity
  • 01:09:44
    n i a
  • 01:09:46
    is known as the magnetic
  • 01:09:48
    dipole moment represented by capital m
  • 01:09:52
    so it's equal to the number of loops
  • 01:09:55
    times the current
  • 01:09:56
    times the area
  • 01:09:57
    that's the magnetic dipole moment
  • 01:10:01
    but this is the equation that we need to
  • 01:10:03
    calculate the torque
  • 01:10:04
    it's
  • 01:10:05
    n i a b sine theta
  • 01:10:09
    now let's say if this is the face of the
  • 01:10:11
    loop
  • 01:10:15
    let's draw the normal line
  • 01:10:17
    perpendicular to the face of the loop
  • 01:10:21
    so this is the area of the loop
  • 01:10:23
    which we could describe it as a
  • 01:10:26
    and the red line
  • 01:10:28
    is the normal line that's perpendicular
  • 01:10:30
    to a
  • 01:10:32
    and sometimes the magnetic field
  • 01:10:34
    it's not going to be
  • 01:10:37
    parallel or perpendicular to the surface
  • 01:10:38
    it can be an angle
  • 01:10:42
    the angle theta is between the magnetic
  • 01:10:45
    field
  • 01:10:47
    and
  • 01:10:48
    the normal line
  • 01:10:49
    which is perpendicular to the surface
  • 01:10:52
    of the coil so make sure you understand
  • 01:10:53
    that
  • 01:10:56
    so theta is between b and the normal
  • 01:10:58
    line
  • 01:11:01
    now instead of drawing a side view of
  • 01:11:03
    the current carrying loop
  • 01:11:06
    let's analyze it
  • 01:11:08
    by drawing the
  • 01:11:09
    top view
  • 01:11:13
    so here's
  • 01:11:16
    this is the top view of it
  • 01:11:27
    and here's the rest of the loop
  • 01:11:30
    if you wish to see it this way
  • 01:11:38
    and this part is perpendicular
  • 01:11:42
    to the plane of the loop
  • 01:11:51
    let's say the magnetic field
  • 01:11:53
    is at a angle of 90 degrees
  • 01:11:59
    when it's at an angle of 90 degrees
  • 01:12:02
    you're going to get
  • 01:12:04
    the maximum torque possible
  • 01:12:09
    so make sure you understand that
  • 01:12:13
    i'm going to redraw like this
  • 01:12:18
    so let's say b
  • 01:12:20
    is directed
  • 01:12:21
    east
  • 01:12:23
    we're going to have one force
  • 01:12:26
    going up and the other force going down
  • 01:12:34
    so this part is perpendicular
  • 01:12:37
    to the area or to the
  • 01:12:40
    the surface of the plane
  • 01:12:42
    of the loop i should say
  • 01:12:45
    and here's the magnetic field
  • 01:12:47
    so it's at an angle of 90 degrees and so
  • 01:12:51
    you're going to get
  • 01:12:52
    the greatest torque in a situation
  • 01:12:56
    that's when the magnetic field is
  • 01:12:57
    perpendicular
  • 01:12:59
    to the normal line which means
  • 01:13:01
    it's parallel
  • 01:13:03
    to the face of the coil
  • 01:13:07
    now let's show another picture
  • 01:13:09
    when
  • 01:13:10
    it's
  • 01:13:12
    at an angle
  • 01:13:15
    so here is the normal line
  • 01:13:17
    which is perpendicular to the face of
  • 01:13:18
    the coil
  • 01:13:20
    and let's draw the magnetic field which
  • 01:13:22
    is directed ease
  • 01:13:24
    so this time
  • 01:13:25
    the angle theta
  • 01:13:27
    is less than 90.
  • 01:13:28
    here theta is equal to 90.
  • 01:13:35
    now f1 and f2
  • 01:13:38
    they're still directed on north and
  • 01:13:40
    south
  • 01:13:41
    but they're much less than the original
  • 01:13:43
    values
  • 01:13:44
    so the torque is less
  • 01:13:47
    and let's see what happens when the
  • 01:13:49
    angle
  • 01:13:50
    is zero
  • 01:13:54
    in this case
  • 01:13:56
    the normal line is parallel
  • 01:13:58
    to the magnetic field
  • 01:14:00
    so therefore the angle
  • 01:14:02
    between the two is zero
  • 01:14:06
    and when this happens
  • 01:14:11
    the magnetic field
  • 01:14:12
    is perpendicular to the face of the coil
  • 01:14:14
    it passes through the coil
  • 01:14:16
    and so there's going to be no torque
  • 01:14:19
    whenever you have this situation
  • 01:14:21
    now let's understand why there's no
  • 01:14:23
    torque
  • 01:14:25
    if you look at the two forces f1 and f2
  • 01:14:30
    notice that they're parallel
  • 01:14:33
    to the level arm anytime you have a
  • 01:14:35
    force that's parallel to the lever arm
  • 01:14:38
    it cannot create a torque
  • 01:14:39
    the only way a torque can be created if
  • 01:14:42
    it's perpendicular let's say this is the
  • 01:14:44
    top view of a door
  • 01:14:47
    here is the level arm r
  • 01:14:49
    the only way for you to push a door
  • 01:14:51
    is to apply a force
  • 01:14:54
    perpendicular to the level arm and the
  • 01:14:56
    door is going to turn
  • 01:14:58
    if you try to push a door from this side
  • 01:15:00
    it's not going to move
  • 01:15:02
    it won't turn
  • 01:15:03
    and that's what's happening here
  • 01:15:05
    the axis of rotation
  • 01:15:07
    is here
  • 01:15:08
    and the way the forces are oriented
  • 01:15:11
    since they're parallel to the level arm
  • 01:15:13
    which is r
  • 01:15:14
    there's gonna be no torque there's no
  • 01:15:16
    rotation
  • 01:15:18
    and so
  • 01:15:19
    that's why whenever the angle is zero
  • 01:15:21
    sine zero is zero so the torque is going
  • 01:15:23
    to be zero
  • 01:15:25
    so the maximum torque occurs whenever
  • 01:15:27
    the angle is 90 that is the angle
  • 01:15:29
    between
  • 01:15:30
    the magnetic field
  • 01:15:32
    and the normal line
  • 01:15:33
    so whenever the magnetic field is
  • 01:15:35
    parallel
  • 01:15:36
    to the face of the coil
  • 01:15:38
    you're going to have
  • 01:15:40
    maximum torque whenever it's parallel to
  • 01:15:42
    the face of the coil it's perpendicular
  • 01:15:44
    to the normal line the angle is 90.
  • 01:15:47
    and whenever the magnetic field passes
  • 01:15:49
    through the coil
  • 01:15:50
    that is when it's parallel to the normal
  • 01:15:52
    line
  • 01:15:53
    there's going to be no torque the angle
  • 01:15:54
    is zero
  • 01:16:00
    now notice what happens
  • 01:16:02
    let's just erase a few things
  • 01:16:08
    i think i'll keep that
  • 01:16:10
    if we look at the first picture on the
  • 01:16:12
    left
  • 01:16:13
    f1 will create a torque that will cause
  • 01:16:16
    it to rotate
  • 01:16:18
    and the clockwise direction
  • 01:16:21
    and f2 will also create a torque
  • 01:16:24
    that will cause the system to rotate in
  • 01:16:25
    the clockwise direction
  • 01:16:29
    now once it rotates it's going to move
  • 01:16:31
    towards a picture 2. it's going to look
  • 01:16:33
    like this
  • 01:16:34
    and as you can see f1 will still create
  • 01:16:36
    a torque
  • 01:16:37
    that will cause the system to rotate
  • 01:16:39
    clockwise and the same is true for f2
  • 01:16:43
    eventually the system is going to reach
  • 01:16:45
    this point
  • 01:16:46
    and
  • 01:16:47
    f1 and f2
  • 01:16:50
    will still be directed north and south
  • 01:16:52
    which we see that in the third diagram
  • 01:16:55
    but once we reach the third diagram
  • 01:16:57
    where the angle is zero
  • 01:17:00
    then the system is at equilibrium
  • 01:17:03
    the two forces
  • 01:17:04
    they're opposite
  • 01:17:06
    and they're equal so they cancel out
  • 01:17:08
    there's no net force in fact there's no
  • 01:17:10
    net force for all pictures
  • 01:17:12
    in the first one these two forces they
  • 01:17:15
    balance each other out
  • 01:17:16
    however there is a net torque for the
  • 01:17:19
    first and second picture but in the
  • 01:17:21
    third diagram
  • 01:17:22
    the net torque is zero
  • 01:17:24
    this force creates no torque and the
  • 01:17:27
    same is true for the second one
  • 01:17:29
    so the net torque and then that force is
  • 01:17:30
    zero for the third diagram which means
  • 01:17:33
    it's at equilibrium
  • 01:17:34
    so basically
  • 01:17:36
    the loop
  • 01:17:38
    moves from this position from an angle
  • 01:17:39
    of 90
  • 01:17:40
    and the magnetic field causes it to
  • 01:17:42
    rotate to an angle of zero and then it
  • 01:17:45
    stops
  • 01:17:46
    now let's work on some problems
  • 01:17:48
    number nine a circular coil of wire has
  • 01:17:51
    a radius of 30 centimeters and contains
  • 01:17:54
    50 loops
  • 01:17:56
    the current is 8 amps and the coil is
  • 01:17:57
    placed in a magnetic field of five tesla
  • 01:18:00
    what is the maximum torque exerted on
  • 01:18:03
    the coil
  • 01:18:04
    by the magnetic field
  • 01:18:05
    so if we draw a picture
  • 01:18:08
    we're gonna have uh
  • 01:18:10
    a coil of wire that is circular and has
  • 01:18:13
    many loops
  • 01:18:15
    50 loops
  • 01:18:17
    and so it's going to look something like
  • 01:18:18
    that
  • 01:18:19
    and the radius
  • 01:18:22
    is 30 centimeters
  • 01:18:26
    to calculate the torque
  • 01:18:28
    it's equal to n i
  • 01:18:31
    a b sine theta
  • 01:18:35
    now if we wish to find the maximum
  • 01:18:36
    torque
  • 01:18:37
    then the angle is going to be 90 and
  • 01:18:40
    sine 90 is equal to 1.
  • 01:18:43
    so n
  • 01:18:45
    is the number of loops and is 50.
  • 01:18:48
    the current
  • 01:18:49
    is 8 amps
  • 01:18:53
    to find the area
  • 01:18:55
    it's going to be the area of a circle
  • 01:18:56
    which is pi r squared
  • 01:18:58
    so it's pi times the radius squared
  • 01:19:00
    the radius is 30 centimeters but we need
  • 01:19:02
    to convert that to meters
  • 01:19:04
    30 centimeters is 0.3 meters
  • 01:19:08
    so it's pi times
  • 01:19:10
    0.3 squared
  • 01:19:15
    so that's the area the magnetic field is
  • 01:19:17
    5 tesla
  • 01:19:20
    and sine 90 is 1.
  • 01:19:25
    so let's start with the area
  • 01:19:27
    0.3 square times pi
  • 01:19:31
    which is 9 pi over 100
  • 01:19:34
    that's 0.2827
  • 01:19:36
    if we multiply that by 5 times 8
  • 01:19:39
    times 50
  • 01:19:42
    you should get 565.5
  • 01:19:47
    tesla
  • 01:19:48
    so don't forget to square 0.3
  • 01:19:52
    and so that's it for this problem
  • 01:19:54
    actually the unit is not tesla that's
  • 01:19:57
    for the magnetic field
  • 01:19:58
    the unit for torque is
  • 01:20:00
    newtons times meters
  • 01:20:04
    so this is it
  • 01:20:07
    number 10
  • 01:20:09
    a rectangular coil contains 200 loops
  • 01:20:12
    and has a current of 15 amps
  • 01:20:14
    what is the magnitude of the magnetic
  • 01:20:16
    field required to produce a maximum
  • 01:20:18
    torque
  • 01:20:19
    of 1200 newtons times meters
  • 01:20:22
    so let's draw a picture
  • 01:20:25
    and so here's a rectangular loop
  • 01:20:27
    and there's 200 loops but i don't want
  • 01:20:29
    to draw it 200 times
  • 01:20:31
    the area is going to be the width
  • 01:20:34
    multiplied by
  • 01:20:36
    the length
  • 01:20:39
    and so we have the dimensions
  • 01:20:42
    so the torque is going to be equal to
  • 01:20:44
    the number of loops
  • 01:20:45
    times the current
  • 01:20:47
    times the area
  • 01:20:48
    times the magnetic field and since we're
  • 01:20:51
    looking for or since we're using the
  • 01:20:52
    maximum torque the maximum torque occurs
  • 01:20:55
    at an angle of 90 degrees
  • 01:20:57
    so this is going to be sine 90.
  • 01:21:01
    now this problem
  • 01:21:02
    the area of a rectangular coil is going
  • 01:21:05
    to be the length times the width
  • 01:21:08
    and sine 90 is 1.
  • 01:21:10
    so the maximum torque is let's put this
  • 01:21:13
    on the left
  • 01:21:14
    it's 1200 newtons times meters
  • 01:21:19
    n
  • 01:21:20
    is equal to 200 loops
  • 01:21:24
    the current
  • 01:21:25
    is 15 amps
  • 01:21:30
    and then the area the length times the
  • 01:21:32
    width
  • 01:21:33
    we need to convert 40 centimeters to
  • 01:21:35
    meters
  • 01:21:36
    so that's 0.4 meters
  • 01:21:38
    times
  • 01:21:39
    0.5
  • 01:21:40
    meters and the magnetic field
  • 01:21:43
    we're looking for b so we got to solve
  • 01:21:45
    for it
  • 01:21:47
    so let's multiply 200 times 15
  • 01:21:50
    times 0.4 times 0.5
  • 01:21:54
    so that's equal to six hundred
  • 01:21:58
    and so now all we need to do is divide
  • 01:21:59
    both sides
  • 01:22:01
    by six hundred
  • 01:22:08
    so these two cancel twelve hundred over
  • 01:22:11
    six hundred we can cancel the two zeros
  • 01:22:13
    so it's twelve divided by six and that's
  • 01:22:15
    two
  • 01:22:17
    so the magnetic field
  • 01:22:18
    is two tesla
  • 01:22:22
    and so that's the answer
  • 01:22:25
    so that's it for this video um if you
  • 01:22:27
    like this video feel free to subscribe
  • 01:22:30
    you can check out my channel for
  • 01:22:32
    more physics videos if you look out for
  • 01:22:34
    my physics video playlist if you want to
  • 01:22:36
    find them all so thanks for watching and
  • 01:22:38
    have a great day
Tags
  • magnetism
  • magnetic poles
  • magnetic fields
  • electric current
  • right-hand rule
  • magnetic force
  • Ampere's Law