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

QuantumCat 2014-09-24 18:50:29	I did this problem in an odd way, but I got to the right answer. Hopefully someone can correct me if there's an error in my reasoning or methodology. So this system is given to be at rest, and that means that the sum of the forces has to be zero. If you equate the centripetal force to the Lorentz force (note the centripetal force is inwardly radial, and if you consider $\vec B$ to be in the $-\hat z$ direction and $\vec v$ to be in the $- \hat \varphi$ direction, the Lorentz force is outwardly radial) you can solve the problem. $m\frac{v^2}{R}$ = $qvB$ $mvR$ = $qBR^2$ $L$ = $qBR^2$ It occurred to me that I should really be using $d/2$ in place of R, but for some reason this works out - I'd suspect that because $B$ only "sees" a rod of radial length $R$ . As I said, I hope someone can correct any error in this methodology, because the approach seems to be too good to be true. Reply to this comment
ramparts 2009-10-01 16:28:18	Once again, limits are your friend. Eliminating D and E by the physical means established in other answers here (that's just stuff you gotta know, or guess), you can consider $R$ and $d$ going to 0 or $\infty$ . As $R$ goes to zero, the magnetic field vanishes, so there should be no angular momentum. Eliminate B. As $d$ goes to $\infty$ , it would make no sense (in fact, it would violate conservation of angular momentum!) if the angular momentum - or the orbital velocity - went to $\infty$ as well, so eliminate C. That leaves A. Reply to this comment
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

buaasyh
2015-10-07 13:29:58

According to Faraday\'s law, E*Pi*d=d Phi/d t. Thus E = (d Phi/d t)/(Pi*d). The force acting on the two pith balls is E*q. Then the corresponding torque is E*q*d/2. Two pith balls contribute twice the torque as E*q*d=q*(d Phi/d t)/Pi. Integrate both sides with respect to t over the process, we have the angular momentum of the system is q times the total change of the magnetic flux over Pi which is q*B*Pi*R^2/Pi=q*B*R^2.

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