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