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GR0177 #86
Problem
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Electromagnetism$\Rightarrow$}Faraday Law

The voltage induced is equal to the change in magnetic flux $\epsilon = -\frac{\partial \Phi}{\partial t}$, where $\Phi =\int \vec{B} \cdot d\vec{A}$.

Noting the initial condition ($\Phi(t=0)=0$), since the field and area normal are perpendicular), one finds that $B \cdot dA = B \sin(\omega t) \pi r^2$. Thus, $d\Phi/dt = \omega B \cos(\omega t) \pi r^2$.

Now, to find the current, one uses Ohm's Law in Faraday's Law to get $IR/N=\dot{\Phi}$, where N is the number of turns. Thus, $I = N/R \dot{\Phi} = NB\omega / R \cos(\omega t) \pi r^2 = 15/2 \times 300/9 \cos(\omega t) \pi (1/100)^2=250E-4\cos(\omega t)$. This is choice (E).

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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 $\theta = 0$ which means $\phi (0) = 0$ and therefore a $\sin$ function.

$\frac{d \phi}{d t}$ 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 $0.025\pi\cos(\omega t)$ as an answer choice. Choice E is $25\pi\cos(\omega t)$. 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 $\Omega$, 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. $\Phi$=NB$\pi$$r^2$sin($\omega$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 $V=IR$ and substitute the expression for the induced voltage given by Faraday's law: $EMF=-\frac{\partial \Phi_B}{\partial t}$ Of course you need to take into account the number of turns...
2006-10-28 21:53:31
You are missing the $\pi$ in your final answer.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$