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GR8677 #81
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Electromagnetism$\Rightarrow$}Faraday's Law

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

where $\Phi=\vec{B}\cdot d\vec{A}$.

The magnetic field at the center of a loop of current-carrying wire is $\vec{B}=\frac{\mu_0 I}{2 r}$, where $r$ is the radius of the loop. (If one forgot this, one can remember it from Ampere's Law.) The magnetic field of the outer loop induces an electric field in the inner loop. The outer loop's B field is $\vec{B}=\frac{\mu_0 I}{2 b}$.

The area of the inner loop stays constant at $\pi a^2$. Thus the flux through it is $\Phi = \frac{\mu_0 I}{2 b} \pi a^2$.

The larger loop carries an ac current, given by $I=I_0 \cos\omega t$. Thus,

$|\frac{d\Phi}{dt}|=\frac{\mu_0 I_0 \omega}{2 b} \pi a^2 \sin\omega t,
$

as in choice (B).

Alternate Solutions
 camarasi2017-10-25 15:31:51 The easiest way is if you remember the equation: $emf = - d\\phi/dt$ which has units $[Bfield] \\times [Area] / [Time]$. \r\n\r\nSince this involves a time derivative, eliminate A and D: the given $cos$ term has to be a $sin$. \r\n\r\nRecognize that the form (excluding numerical factors) $\\mu_0 I_0 /R$ has units of B-field - think of the field of a current loop. Mentally extract a (b) term from the denominators of answers B, C, and E into the parenthesis and consider it a B-field. Now you need to look for the $[Area]\\[time]$ term.\r\n\r\nEliminate E now: no time dependence.\r\n\r\nEliminate C: No $[Area]$ term.\r\n\r\nB is the correct answer: both area term $a^2$ and time term $1/\\omega$ exist. \r\n\r\n\r\n\r\n\r\nReply to this comment ramparts2009-11-06 10:55:30 Here's yet another quick GRE-type approach (if you're good with units) - remembering or deriving that a weber is a volt-second, and using the definition of $\mu_0$ from the front of the test, you can rule out all but A and B based on units. Now, if $\omega=0$ then there's a non-zero but constant current in the outer loop. This isn't going to induce anything, so pick the one with the sine term, as that goes to 0 in that limit.Reply to this comment wittensdog2009-07-27 12:13:32 The way I approached it was to remember the two things which seem to be what ETS is looking for here. One is that there is a time derivative, which should create a term of the form w * sin (wt). That eliminates A, D, and E, the first two because they have cosine instead of sine, and the last one because there is no w that comes out front out of the cos(wt) as a result of the chain rule. The second is that since the inner loop is said to be very small compared to the larger loop, you can take the field to be roughly the same over the inner loop, and so the flux should be proportional to the area of the inner loop, which goes with a^2. All of the parenthetical terms are the same, so this leaves only B. Then I guess for free you get the fact that there should be a single factor of b in the bottom. That analysis seems faster to me than unit checking.Reply to this comment dicerandom2006-09-06 20:46:32 There's another way to think through this which doesn't involve having to remember (or derive) the magnetic field of the loop. We know that the field generated by the loop will have a cos(wt) dependence since the current has that dependence. Thus the EMF must have a sin(wt) dependence and we can eliminate the options involving cos(wt). Of the remaining options only (B) has the proper units (mu0 ~ gauss/(meter*ampere) ).Reply to this comment
casseverhart13
2019-08-15 08:50:40
Good Problem Examine and fascinating. refrigerator repair
camarasi
2017-10-25 15:31:51
The easiest way is if you remember the equation: $emf = - d\\phi/dt$ which has units $[Bfield] \\times [Area] / [Time]$. \r\n\r\nSince this involves a time derivative, eliminate A and D: the given $cos$ term has to be a $sin$. \r\n\r\nRecognize that the form (excluding numerical factors) $\\mu_0 I_0 /R$ has units of B-field - think of the field of a current loop. Mentally extract a (b) term from the denominators of answers B, C, and E into the parenthesis and consider it a B-field. Now you need to look for the $[Area]\\[time]$ term.\r\n\r\nEliminate E now: no time dependence.\r\n\r\nEliminate C: No $[Area]$ term.\r\n\r\nB is the correct answer: both area term $a^2$ and time term $1/\\omega$ exist. \r\n\r\n\r\n\r\n\r\n
ramparts
2009-11-06 10:55:30
Here's yet another quick GRE-type approach (if you're good with units) - remembering or deriving that a weber is a volt-second, and using the definition of $\mu_0$ from the front of the test, you can rule out all but A and B based on units. Now, if $\omega=0$ then there's a non-zero but constant current in the outer loop. This isn't going to induce anything, so pick the one with the sine term, as that goes to 0 in that limit.
 natestree2011-09-22 11:39:38 But B is the answer...
 michael2011-10-08 15:04:37 This doesn't work. If $\omega = 0$ both (a) and (b) go to zero.
wittensdog
2009-07-27 12:13:32
The way I approached it was to remember the two things which seem to be what ETS is looking for here. One is that there is a time derivative, which should create a term of the form w * sin (wt). That eliminates A, D, and E, the first two because they have cosine instead of sine, and the last one because there is no w that comes out front out of the cos(wt) as a result of the chain rule. The second is that since the inner loop is said to be very small compared to the larger loop, you can take the field to be roughly the same over the inner loop, and so the flux should be proportional to the area of the inner loop, which goes with a^2. All of the parenthetical terms are the same, so this leaves only B. Then I guess for free you get the fact that there should be a single factor of b in the bottom. That analysis seems faster to me than unit checking.
 flyboy6212010-11-15 19:23:19 Yes
 lelandr2011-04-24 22:41:47 +1 for this solution
wangjj0120
2008-08-28 07:55:38
I think $\Phi = \frac{\mu_0 I}{2 b} \pi a^2$ is incorrect because $\frac{\mu_0 I}{2 b}$ is the magnetic field only at the center. Magnetic field induced by the outer current should be a function of r, that is B=B(r). To get the magnetic flux, we should integrate $\Phi=\int\vec{B}(r)\cdot d\vec{A}$ instead of $\Phi = \frac{\mu_0 I}{2 b} \pi a^2$.
 gt20092009-06-25 16:38:29 It said approximately equal to.
ivalmian
2008-04-03 21:54:38
Just a little notice - you can't use Ampere Law for this problem, instead use Biot-Savart.
 syreen2013-09-12 19:15:46 Why not? I can't figure out why, but Amperes gives me an e0 instead of pi and switches b and a.
dicerandom
2006-09-06 20:46:32
There's another way to think through this which doesn't involve having to remember (or derive) the magnetic field of the loop.

We know that the field generated by the loop will have a cos(wt) dependence since the current has that dependence. Thus the EMF must have a sin(wt) dependence and we can eliminate the options involving cos(wt). Of the remaining options only (B) has the proper units (mu0 ~ gauss/(meter*ampere) ).
imrebartos
2005-11-09 07:11:15

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