A rectangular loop, dimensions H, W falls from a region...

A rectangular loop, dimensions H, W falls from a region where the uniform magnetic field of a given magnitude B0 points out of the page. The loop has resistance R. a) Find the current (magnitude and direction) in the loop as a function of the velocity of the loop v once the lower segment leaves the region with magnetic field. (Note that the positive y direction is down). Ignore the self-inductance. b) Find the magnetic force acting on the frame as a function of the loop's velocity when the frame is (i) completely within the region of the magnetic field; (ii) half way falling out of the magnetic field, and (iii) out of the region of the magnetic field.

Transcript

00:01 All right, here we have a red loop of wire.

00:04 It is falling through a magnetic field that is pointing towards us.

00:09 So at first, before it gets to the edge, there's no change in flux within the box.

00:16 And so there's no current flowing in our loop.

00:20 But when we hit this point here, so i'll kind of show maybe i'll change color, make it black.

00:25 So when we are right here where we're half in and half out or any portion out, but not all the way out.

00:31 At this point now, we are changing the flux.

00:35 So what's happening is our flux towards us is decreasing.

00:39 The loop will create a current in such a way is to create a magnetic field which opposes the change of flux.

00:44 So we want more out.

00:47 So we will actually create, or the loop, loop will create a counterclockwise current.

00:52 So the current will be counterclockwise.

00:55 As we are exiting the loop, it'll try to make more be felt out to reduce the change.

01:00 Okay, so our first step is to solve for the magnitude of that counterclockwise current.

01:07 Okay, so it will be equal to our biomes law, emf over our voltage over resistance, and we're told that we do have a resistance of the loop.

01:17 And the emf is minus a rate of change of flux with time over r then.

01:24 So faraday's law tells me that the emf has been the rate of change of flux.

01:28 The b is not changing, but the area is changing, and that's what we need to solve for and then over r.

01:35 So let's solve for the kind of the area and what's going on with the area below.

01:40 Okay, so the area then in general is if i have the full area, it would be height times width.

01:47 But as i'm exiting, then i still get the width, but my height is being reduced by the fact that i'm speeding out of there.

01:57 So distances rate times time, so the amount i have exited will be b times t.

02:04 So my d -a -d -t then will be equal to w times the derivative of the inside, which is just b minus b.

02:12 So i get minus w -w.

02:14 Whoa, that should be minus wp.

02:17 So that sticks up or minus wv.

02:21 Okay, so now i know that my clockwise current equals to minus b a double negative and types wv over r.

02:32 So that is my clock wide a counterclockwise current...

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