0:00
Hi there.
00:01
So for this problem, we have a coil of wire consisting of 40 rectangular loops.
00:08
So the number of loops in this case is equal to 40.
00:13
And with a width that is equal to 16 centimeters.
00:21
So the width is equal to 16 centimeters.
00:27
And a height, which is equal to 30 centimeters.
00:33
So it's placed in a constant magnetic field that is given in a vector form is equal to 0 .65 tesla in the s direction and plus 0 .250 tesla in the y direction.
01:09
Seat derruption, sorry, here, seat, see direction.
01:16
So the coil is hinged to a fitzit think rod along the y -axis, and it is originally located in the s -y plane.
01:31
So there is a current equal to 0 .2 ampers that runs through the wire.
01:44
So for part a of this problem, we need to determine what are the magnitude and the direction of the force fab that b observes on the sediment ab of the coil.
02:00
So the situation that we have for this problem is shown in this figure right here.
02:10
As you can see, the width in here 16, the height is 30 centimeters.
02:15
It is located in the xy plane and it is rotated in here as is shown in the figure.
02:24
So we know that the force on a length of wire carrying current in the bed tour form is equal to the number of thorns times and the current times the length the vector cross -product with the magnetic field.
02:52
And then the force on the settlement ab, the settlement right here is 1.
03:02
Ab is going to be the following in the vector form is equal to n times the end dorms, times the magnitude for the current, times the length l -a -b times, well, this is, of course, in this case, the current is traveling in the x direction.
03:37
That's why we have the unit vector in there.
03:40
And this cross -product with the magnetic field.
03:49
And we know that the magnetic fuel has two components.
03:56
So we have the ads component and the seat component.
04:03
So as you can see, we are going to have the cross product between ads and ads and the cross product between ads and theta.
04:14
We know that the cross product between the same unit vector is equal to.
04:20
To zero and the other one it will give us my it will give us yes minus y so from this we simplify this expression to be minus the number of forms times the current and in this case the length of that segment as you can see corresponds to the width so that is the current times the width the seed component of the magnetic field and this in the y component.
05:06
So in here we just need to simply substitute all of these values in order to obtain this better.
05:14
So in this case we know that the number of forms is 40.
05:22
The current is 0 .2 ampers.
05:25
The width.
05:25
Is 0 .16 meters.
05:29
The magnitude for the magnetic field is 0 .250 tesla, and all of this in the y component.
05:39
And this is minus 0 .32 in the y direction.
05:46
And of course, in the units of neutums.
05:49
So that's the solution for part a of this problem.
05:53
Now, for part b, we need to determine what are the magnitude and the direction of the force fbc that the magnetic field exerts on a sediment and bc on the coil.
06:10
So in the case b we need to determine the petur of the force bc in that segment.
06:19
As you can see, that semic corresponds to here.
06:24
And so with that said, in this case, we are going to have that that vector is equal to the number of terms times the current through that segment times the length bc that we know that in this case corresponds to the height.
06:46
And this times, well, this is seen in the y direction, as you can see.
06:53
The current is now traveling in the y direction, in the positive y direction.
06:58
And this, the cross -product to the magnetic field, which again has two components...