B-field to create average emf in rotating coil

AI Thread Summary
The discussion revolves around calculating the magnetic field strength required to induce an average EMF of 10,000 V in a rotating coil with specific parameters. Initial attempts to solve the problem using the formula E = BANω resulted in incorrect values for B, leading to frustration. Participants emphasize the importance of using the correct equations and understanding the average EMF over time, suggesting a need for integration rather than simple calculations. One contributor expresses disappointment in the lack of helpful responses from the forum, highlighting a broader concern about the reliability of online assistance for physics homework. The conversation underscores the challenges of conveying complex physics concepts effectively.
ChuckB
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Homework Statement


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.[/B]

Homework Equations


E = BANomega
solving for B, so rewrite as B= E/(ANomega)[/B]

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
 
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ChuckB said:
E = BANomega
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.
 
Qwertywerty said:
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.
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?
 
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haruspex said:
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?
I believe a linear relation between a variable, and time would be required to simply use the initial and final condition.
 
Qwertywerty said:
I believe a linear relation between a variable, and time would be required to simply use the initial and final condition.
Why? I don't need that to find average velocity or average acceleration. ##\vec v_{avg}=\Delta \vec s/\Delta t##.
 
Hmm..After working out a problem, I've realized that I've made a mistake. Thanks for correcting me @haruspex.
 
60 rev/s = 120(pi) rad/s, not 120 rad/s as I mistakenly stated and used in my calculations.
 
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
 
ChuckB said:
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
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:
haruspex said:
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?
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
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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  • #11
ChuckB said:
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
I suggest both are useful.
Making sure you have understood what others tell you is also rather helpful.
 
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  • #12
The first principle of advice we give to others is that it applies equally to ourselves.
 

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