GR8677 #81



Alternate Solutions 
camarasi 20171025 15:31:51  The easiest way is if you remember the equation: which has units . \r\n\r\nSince this involves a time derivative, eliminate A and D: the given term has to be a . \r\n\r\nRecognize that the form (excluding numerical factors) has units of Bfield  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 Bfield. Now you need to look for the term.\r\n\r\nEliminate E now: no time dependence.\r\n\r\nEliminate C: No term.\r\n\r\nB is the correct answer: both area term and time term exist. \r\n\r\n\r\n\r\n\r\n   ramparts 20091106 10:55:30  Here's yet another quick GREtype approach (if you're good with units)  remembering or deriving that a weber is a voltsecond, 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 nonzero 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 20090727 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 20060906 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) ).  

Comments 
casseverhart13 20190815 08:50:40  Good Problem Examine and fascinating. refrigerator repair   camarasi 20171025 15:31:51  The easiest way is if you remember the equation: which has units . \r\n\r\nSince this involves a time derivative, eliminate A and D: the given term has to be a . \r\n\r\nRecognize that the form (excluding numerical factors) has units of Bfield  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 Bfield. Now you need to look for the term.\r\n\r\nEliminate E now: no time dependence.\r\n\r\nEliminate C: No term.\r\n\r\nB is the correct answer: both area term and time term exist. \r\n\r\n\r\n\r\n\r\n   ramparts 20091106 10:55:30  Here's yet another quick GREtype approach (if you're good with units)  remembering or deriving that a weber is a voltsecond, 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 nonzero 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.
natestree 20110922 11:39:38 
But B is the answer...

michael 20111008 15:04:37 
This doesn't work. If both (a) and (b) go to zero.

  wittensdog 20090727 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.
flyboy621 20101115 19:23:19 
Yes

lelandr 20110424 22:41:47 
+1 for this solution

  wangjj0120 20080828 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 instead of .
gt2009 20090625 16:38:29 
It said approximately equal to.

  ivalmian 20080403 21:54:38  Just a little notice  you can't use Ampere Law for this problem, instead use BiotSavart.
syreen 20130912 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 20060906 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 20051109 07:11:15   




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