# B-field to create average emf in rotating coil

1. Mar 13, 2016

### ChuckB

1. The problem statement, all variables and given/known data
A 0.300 m radius, 487 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.

2. Relevant equations
E = BANomega
solving for B, so rewrite as B= E/(ANomega)

3. The attempt at a solution
r = 0.3m
A = 0.28274 m^2
t = .004175s
N = 487
theta (dif) = 90 deg
theta (start) = 0 deg
theta (final) = 90 deg
omega = 60 rev/s = 120 rad/sec

Vavg = Vmax * .637
so 10000/.637 = 15699V

substitute values: B = 15699/(0.28274*487*120) = 0.95T X incorrect!
perhaps E = 10000?
substitute values: B =10000/ (0.28274*487*120) = 0.605T X incorrect!

After trying other approaches that involved deltaT, this I felt was the best chance I had at getting the correct answer. Now I am stumped and frustrated. Any/all help will be very greatly appreciated.

ChuckB

2. Mar 13, 2016

### Qwertywerty

Which formula is this?
What is the E.M.F induced at any time $t$, if the angle betwee the plane of the coil, and the magnetic field is $θ$?

For any variable, what is it's time average? $\frac {∫x.dt} {∫dt}$ You must use this formula right here, and not the one you used, for the average value of E.M.F.

3. Mar 13, 2016

### haruspex

Is it really necessary to use an integral? Rewriting the quoted form of Faraday's Law as $E=-N\frac{\Delta (BA)}{\Delta t}$, is it not sufficient to consider the change in area in a suitable time interval?

4. Mar 13, 2016

### Qwertywerty

I believe a linear relation between a variable, and time would be required to simply use the initial and final condition.

5. Mar 13, 2016

### haruspex

Why? I don't need that to find average velocity or average acceleration. $\vec v_{avg}=\Delta \vec s/\Delta t$.

6. Mar 13, 2016

### Qwertywerty

Hmm..After working out a problem, I've realised that I've made a mistake. Thanks for correcting me @haruspex.

7. Mar 13, 2016

### ChuckB

60 rev/s = 120(pi) rad/s, not 120 rad/s as I mistakenly stated and used in my calculations.

8. Mar 13, 2016

### ChuckB

I cannot overstate how disappointed I am in this forum as a source for useful assistance with physics homework. After wading through the SLUDGE at Chegg and Yahoo Answers, finding nothing but half-baked, slipshod, and generally simply WRONG advice, I thought that a dedicated physics forum with actual physicists on the premises might be a God-send. Instead, and although I'm stumped by it, what I recognize as a fairly straightforward General Physics II question, with all the information necessary presented in a clear manner, has resulted in ZERO useful assistance.

I am resigning myself to the fact that the public Internet will inevitably tend toward entropy and that any advice received on the Internet is automatically suspect.

Sincerely and with regrets,
ChuckB

9. Mar 13, 2016

### haruspex

Like it or not, the principle of the homework forums on this site is to provide hints and corrections. We do not do your thinking for you.
Perhaps my hint was too subtle:
That formula gives you the average field over time interval Δt. You want the long term average, so you need to pick a representative interval.

10. Mar 13, 2016

### ChuckB

You have something valuable to say, however I'll tell you the same thing I tell my physics professor: success in life doesn't hinge on what you know, but rather on your ability to convey and explain it to people who aren't anything at all like you. Your message was fine, but I am uninspired to filter out the backscatter and interference to get to a potentially valuable nugget. Attached is the completed and verified correct solution to the problem.

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• ###### Phys_Ch23_Q4.pdf
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11. Mar 13, 2016

### haruspex

I suggest both are useful.
Making sure you have understood what others tell you is also rather helpful.

Last edited: Mar 13, 2016
12. Mar 13, 2016

### ChuckB

The first principle of advice we give to others is that it applies equally to ourselves.