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GR0177 #86
Problem
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The voltage induced is equal to the change in magnetic flux , where .

Noting the initial condition (), since the field and area normal are perpendicular), one finds that . Thus, .

Now, to find the current, one uses Ohm's Law in Faraday's Law to get , where N is the number of turns. Thus, . This is choice (E).  Alternate Solutions
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Copperknickers
2016-07-08 06:46:05
Is there a simpler way to figure out the cos(wt), sin(wt) part? Without having to remember any formula by rote, that is? asdfman\'s solutions seems nice. Any alternative ones?
 asdfasdfasdf2016-09-07 14:53:11 I\'m not sure there\'s a simpler way to figure out the trig part. The only \"memorization\" needed is: (a) what are sin(0) and cos(0)? (b) what is the derivative of sin(wt) or cos(wt)? The question provides the initial condition of the n vector being parallel to the y-axis. When that is the case, there is no magnetic flux through the coil, right? So, when t=0, we should have B=0. sin(0)=0, so we\'ll use sine, not cosine. And you know you need to use some sort of periodic function since it\'s rotating...once it rotates such that the n vector lies parallel to the x-axis, we have the maximum flux possible, right? sin(90)=1, so that confirms that we should use sine.Then, since we get our emf from d(phi)/dt, we need the derivative of sin(wt), which is w*cos(wt).Does that help? [also, sorry if this text gets garbled...I\'m not sure why slashes are being randomly inserted.] John Smith
2013-10-17 02:02:09
I am a foreigner, the time when I solve this question, I can hardly notice that the FUNK MILLIamperes. That will cost me lots of time if it happened in the real test. asdfman
2009-11-02 23:47:03
it says the coil starts at which means and therefore a function.

must therefore be a cosine.

Do some algebra and you realize that it'll be something like 1.5^2, so it'll have something ~2.25 or 2.5. The only answer that fits the bill is E. theevilmachines
2009-07-12 17:58:53
I don't know about you guys, but my 0177 test doesn't have as an answer choice. Choice E is . Am I missing something?
 theevilmachines2009-08-24 15:31:09 Oh it says in milliamperes, not amperes. duckduck_85
2008-11-03 22:20:12
how do you know that the resistance given in the problem isn't for the entire coil? (so that you needn't take the # of turns into account)
 elzoido2382008-11-07 19:17:30 The problem states that "...the coil resistance is 9 , which I interpret as the resistance of the entire coil. Since Yosun found the flux for 1 turn of the coil, you would need to account for the number of turns in the coil by dividing the total resistance by N (if, however one found the flux for the entire coil - i.e. =NBsin(t) - one would not need to find the resistance per turn.) Thanks Yosun for this kick-ass site! :) sblmstyl
2008-10-15 21:39:49
thanks yosun! naama99
2006-11-29 21:16:46
What do you mean by \"one uses Ohm's Law in Faraday's Law"? Can you be more specific about it? It seems to me that you are taking the R given in the question and merely using it as L
 Richard2007-10-29 09:53:54 I'm pretty sure she just means, take Ohm's law and substitute the expression for the induced voltage given by Faraday's law: Of course you need to take into account the number of turns... 2006-10-28 21:53:31
You are missing the in your final answer.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$