In Fig. P29.51 the loop is being pulled to the right at...

In $\textbf{Fig. P29.51}$ the loop is being pulled to the right at constant speed $v$. A constant current $I$ flows in the long wire, in the direction shown. (a) Calculate the magnitude of the net emf $\varepsilon$ induced in the loop. Do this two ways: (i) by using Faraday's law of induction ($Hint:$ See Exercise 29.7) and (ii) by looking at the emf induced in each segment of the loop due to its motion. (b) Find the direction (clockwise or counterclockwise) of the current induced in the loop. Do this two ways: (i) using Lenz's law and (ii) using the magnetic force on charges in the loop. (c) Check your answer for the emf in part (a) in the following special cases to see whether it is physically reasonable: (i) The loop is stationary; (ii) the loop is very thin, so $a \rightarrow$ 0; (iii) the loop gets very far from the wire.

Transcript

00:02 Okay, so we're going to do chapter 29, problem 51.

00:06 So we have this picture over here, and what this has is we have a current carrying wire here with the current going upwards.

00:16 And next to this, there's a wire loop of dimensions a by b and a square rectangular loop, and its position r away from the wire.

00:28 And it says it's moving to the right and the positive x direction to the right away from the wire at a speed v.

00:36 Okay, so first it wants us to calculate the emf induced on this wire a couple different ways.

00:48 Now, usually this is pretty easy because it's just faraday's law is given as the magnitude is given by the change in magnetic flux in perspective of time.

00:57 But this can be hard if the flux is not constant.

01:01 And that's exactly the case we have here.

01:03 Because some parts of the loop are farther away from the wire or closer, and the b field from the wire is dependent on the distance away.

01:15 It's dependent on b of r.

01:17 Okay.

01:18 So let's write down what the d field of a wire is.

01:22 So this comes from a previous example here.

01:28 So it's u not times the curve.

01:30 Through the wire over 2 pi and new distance away.

01:34 We're going to be in the x direction here.

01:36 So we have this.

01:39 So now we need to write what the differential of flux is going to be.

01:44 So this is as a is our constant, but b now changes with x.

01:48 So this is db.

01:51 And this is the same thing as b times this will be times d x.

02:02 Okay.

02:04 So now we can write this.

02:05 This out again to learn what b is here.

02:11 So we're off with m0, i, and b over 2 pi, and this is dx over x.

02:21 Cool.

02:22 So now we can finish this out to figure out what the total flux is going to be.

02:30 We're off to integrate this.

02:31 So we can see that the total flux is given as the integral of the dribble flux.

02:39 And we can rewrite this now as what we, our expression for the derivative is, g over 2 pi.

02:52 And now we have an integral.

02:54 And since our differential here is dx, so we're integrating from x.

02:59 So our x starts at r, because that's for the first, and it adds up a.

03:03 So we're going to integrate from r to r plus a.

03:08 That's our bounds, and the integral is dx over x.

03:14 So we can just do that integral now, and we see that this becomes mu not i b over 2 pi times the natural log of r plus a over r.

03:33 So this is what we have magnetic flux.

03:38 But we see for faraday's law, we need the time derivative of this.

03:43 So we got to figure out what this is.

03:48 This is the same thing as the r derivative in the chain rule here, which is the same thing as the time derivative of flux times the speed.

04:04 Cool.

04:05 So we can plug all that in and we should see that the magnitude of our emf should be the magnitude of r derivative.

04:17 That's r derivative.

04:18 R derivative.

04:19 So, mu not i b over 2 pi.

04:26 Then we take the v, now take the r derivative of this natural log and we should be left with negative a over r, r plus a.

04:39 So we simplify this all out, and we're left with mu not i, a b, v over 2 pi r times r plus a.

04:53 Cool.

04:55 We got there.

04:56 So now we want to check this by instead of using faraday's law and integrating.

05:03 Let's do this another way.

05:05 Another way we can do it is find emotional emf from bvl for each bar in our circuit.

05:15 So if this is our circuit, we can label these as one, two, three, and four.

05:22 Okay.

05:24 And we can easily see here that the total is just going to be the sum.

05:29 All of these.

05:32 So from i is from one to four...

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