GREPhysics.NET: GR9677 #87

GR9677 #87

Problem

GREPhysics.NET Official Solution

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Electromagnetism $\Rightarrow$ }Faraday Law

From Faraday's Law or Len's Law, one has $\vec{E}\cdot d\vec{l} = -d\Phi/dt$ . Since, in order for the balls to move, they must move in a circle, one has $dl=2\pi d/2$ ; moreover, the induced magnetic field would point in the opposite direction to the field that was before, and one has a current in a loop from the right-hand-rule. The area of the magnetic flux is just $\pi R^2$ , since the field only goes through the cylindrical region of radius R.

Thus, $E(2\pi d/2 ) = \dot{B}\pi R^2 \Rightarrow E=\frac{\dot{B}R^2}{d}$ .

Now, recall some mechanics. The torque is related to the moment-arm and force by $\tau = \sum \vec{r} \times \vec{F}$ , where $\vec{F}=q\vec{E}=q\frac{\dot{B}R^2}{d}$ . Since there is a force contribution from each charge, and since, by the right-hand-rule, their cross-products with the moment-arm point in the same direction, one finds the torque to be $\tau=2(d/2)q\frac{\dot{B}R^2}{d} = dq\frac{\dot{B}R^2}{d}=q\dot{B}R^2$ .

Now, recall the relation between angular momentum and torque to be $\sum \tau = \dot{L}$ . Replace the $\dot{B}\rightarrow B$ above to get $L = qBR^2$ , and so the system starts rotating with angular momentum as in choice (A). (This approach is due to Matt Krems.)

Note that one can immediately eliminate choice (D) since angular momentum is not conserved from the external torque induced (to wit: electromagnetic induction). Moreover, although choice (E) is true in general, it does not apply to this problem.

Alternate Solutions

student2008 2008-10-17 09:36:12	The electromagnetic field had the moment assosiated with the Umov-Pointing vector $\vec S \sim [\vec E,\vec B]$ , where $\vec E$ was the electorstatic field of the charges. When the magnetic field is turned off, the field moment should conserve and thus transfer into the mechanical momentum of our system. Then, one may notice that as $d\to\infty$ , $E_{initial}\to 0$ , so $L\to 0$ or at least $L$ is limited. Only the answer (A) works this way. Reply to this comment
shoyer 2007-08-05 23:27:36	There is an easier way to solve this problem. First, you can eliminate E because even though magnetic forces do no work, that doesn't mean they can't cause systems to move. The rod rotating would not be in a higher energy state. Answers A-C assume there was angular momentum stored in the field -- answer D claims otherwise. I wasn't entirely sure about eliminating D but it didn't feel right (I wasn't sure about the induced force but that explanation in the official answer makes sense). Unlike the official answer, I'm considering the momentum of the entire system. In that case, since the magnetic field's size has nothing to do with the positions of the balls, the stored momentum in that field should not be dependent on d. This eliminates B and C, leaving A as the answer. Reply to this comment

Comments

student2008
2008-10-17 09:36:12

The electromagnetic field had the moment assosiated with the Umov-Pointing vector $\vec S \sim [\vec E,\vec B]$ , where $\vec E$ was the electorstatic field of the charges. When the magnetic field is turned off, the field moment should conserve and thus transfer into the mechanical momentum of our system. Then, one may notice that as $d\to\infty$ , $E_{initial}\to 0$ , so $L\to 0$ or at least $L$ is limited. Only the answer (A) works this way.

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casaubon
2008-10-04 18:19:31

where does this dl come from in Faraday's law?rnrnEverything I see just has $E=-d\Phi\dt$ , why is it $E\cdot dl$ now?

casaubon
2008-10-15 18:34:57

so to address my own question,

$V=-\frac{d\Phi}{dt}$

$E\cdot dl=V$

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p3ace
2008-05-05 01:36:02

This is an interesting problem. Lenz's Law is easy to brush off as just explaining the minus sign in Faraday's law, but it really should be given more emphasis. One cannot determine the direction without understanding what the minus sign really means. If the change in B is an increase in B then the induced field opposes the original direction of B, but if the change in B is a decrease, then the direction is IN THE SAME DIRECTION AS B.

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p3ace
2008-04-11 13:49:21

To answer my own question, I suppose that it is the induced electric field that actually does the work, and not the magnetic field that did the inducing, although, the work wouldn't get done with out the change in the B field. I suppose that what ever caused the change in the B field did the original work.

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p3ace
2008-04-11 13:44:37

Also, if magnetic forces do no work, then why is W=(1/2)Li^2?

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p3ace
2008-04-11 13:33:24

Since the magnitude is all that is asked for in this problem, it doesn't matter, but the direction doesn't oppose the direction of the original B field. It opposes the change, whatever direction that is. If the change in B is a decrease, the change is in the opposite direction of B. Thus, the induced B field is in the same direction as the original B, which is opposite the change.
Another thing is, at first I was thinking about a current being induced with the axis of rotation downward so that the charges circulate from right to left on the near side. This is one way to look at it. When I think of angular momentum, I think of it having mass, and in this case it doesn't, which is a little confusing.

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bkardon
2007-09-30 09:44:17

This is just a tripped up magnetic induction problem, like problem #2. When you take away the magnetic flux, you will induce a current in the "pith" balls; i.e. you will make the rotate. Without doing much work, you know that the voltage is proportional to the change in flux, which is dependent on R^2. We can also safely assume that the angular momentum is not related to d, since the angular momentum induced shouldn't be dependent on the arrangment of charges carrying it. This indicates choice A.

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shoyer
2007-08-05 23:27:36

There is an easier way to solve this problem.

First, you can eliminate E because even though magnetic forces do no work, that doesn't mean they can't cause systems to move. The rod rotating would not be in a higher energy state.

Answers A-C assume there was angular momentum stored in the field -- answer D claims otherwise. I wasn't entirely sure about eliminating D but it didn't feel right (I wasn't sure about the induced force but that explanation in the official answer makes sense).

Unlike the official answer, I'm considering the momentum of the entire system. In that case, since the magnetic field's size has nothing to do with the positions of the balls, the stored momentum in that field should not be dependent on d. This eliminates B and C, leaving A as the answer.

evanb
2008-07-02 13:50:36

See Griffith's page 358.

However, the stored momentum is dependent on $\vec{E} \cross \vec{B}$ , so the stored momentum could be dependent on d, through $\vec{E}$ .

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vinograd19
2007-02-02 10:00:05

There is no fields around balls. Thus there is no forces to move it.

lukassvec
2007-07-25 00:21:28

I though so to in the beginning but then again, there is an electric field inside the magnetic field.

newton
2007-09-27 21:19:17

a changing magnetic field creates an electric field-Faradays law

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