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GR8677 #81
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Alternate Solutions |
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  from the front of the test, you can rule out all but A and B based on units. Now, if  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. |  | 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. | | 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) ). | |
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Comments |
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 from the front of the test, you can rule out all but A and B based on units. Now, if 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. | | 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. | | wangjj0120
2008-08-28 07:55:38 | I think  is incorrect because  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 \cdot d\vec{A}) instead of .
gt2009
2009-06-25 16:38:29 |
It said approximately equal to.
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|  | ivalmian
2008-04-03 21:54:38 | Just a little notice - you can't use Ampere Law for this problem, instead use Biot-Savart. | | 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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