# Faraday's Law of Induction Question

1. May 18, 2004

### paul11273

I can't seem to wrap my head around this question. Any help would be great.

A coil consists of 20 turns, with a 10cm radius (all turns are uniform).
It rotates uniformly around the z-axis at an angular speed of: w=20 rad/s
A uniform magnetic field passes along the x-axis with a magnitude of B=0.3T.

Find:
a) The induced emf at any time moment t>0 if at t=0 the coil was in the yz plane.
b) Direction of the induced current at t=0.

First, b) is easy. The answer is zero, there is no induced current at t=0 because the coil is not moving through the field yet.

For a) I started with this:
E=-N*(dB flux/ dt) (where E=emf)

Since B flux=BAcos(theta)=BAcos(wt) is sub this in to get:

E = -NBAcos(wt) d/dt
E = NBAwsin(wt) then sub in the given values for N, B, w and calculate A=pi*r^2

E = 20*.3*pi(.1m)^2*20*sin(20t)
E = 1.2pi*sin(20t) is my final answer.

Is this correct? Someone in my class has E=4pi*sin(20t)

Can someone check my answer and offer some advice?
Thanks in advance.

2. May 18, 2004

### AKG

This doesn't sound right. In fact, there is no reason to assume that the coil isn't moving at t=0. t=0 was just arbitrarily chosen to be some point in time when the coil was in a specific orientation. Only if the question stated that the coil was rotating for t > 0 would your answer make sense. Otherwise you should assume that it has always been rotating.
This looks right to me. Have you seen the other person's work? Maybe you can figure out why they got a "4" instead of "1.2".

3. May 18, 2004

### paul11273

OK, I understand your point. I could be wrongly assuming that the coil is stationary at t=0. But, even if the coil was in motion prior to t=0, and we are defining the coil location at t=0 to be when the coil is right on the yz plane, then wouldn't the current still be zero? This would be the exact point where the current reverses direction.

What do you think?

4. May 19, 2004

### AKG

Perhaps you know something I don't. If you're given a resistance of the coil, you can find the current as:

$$i(t)\ =\ v(t)/R$$
Maybe you know of some formula that tells you the current induced in a coil when it's moving in a magnetic field. Well, you know how to find the EMF. My textbook has the following formula:

$$V\ =\ -L\frac{dI}{dt}$$ (*)

You might want to try rearranging and integrating so that you have "I = ...." Here, they talk about a solenoid which has current "I" passing through it, and then give a way to find "V," I'm not sure, but maybe you can start with "V" and use that to determine the current passing through it. The only thing is, when you pass a current through it, that creates a magnetic field which in turn creates an induced EMF, whereas you use the formula:

$$V\ =\ -\frac{d\Phi_{B}}{dt}$$

when you already have a magnetic field moving relative to the coil. I'm not sure if that difference is enough to prevent you from using the formula (*). Anyways, there's that, that's my best suggestion. Also, if you don't have it, there is the formula:

$$L\ =\ \mu_{o}N^2\frac{A}{l}$$

If you need to find the inductance of the solenoid (inductor). Actually, from what I see, you're not given "l," but like I said, maybe you know more than I do.

5. May 19, 2004

### Staff: Mentor

I also assumed (as did AKG) that the coil was always rotating at a uniform angular speed. (If not, then it would have infinite acceleration at t = 0!) But, nonetheless, you are correct that the EMF (and thus the induced current) is zero at the point where t=0.

You already found that the induced EMF goes as sinωt. Thus at t=0, the induced EMF is zero.

6. May 19, 2004

### paul11273

No, the resistance was not given. The question only asked us to sketch the direction of the current at t=0. I think this was sort of a trick by the prof. to see if we have the right concept. As Doc Al pointed out, if my part a) answer is correct then subbing in 0 for t makes the current 0 at that time.

Thanks to both of you for your help. I feel pretty confident on this question now, after having gone through it with both of you.

Thanks again.

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