1. Oct 21, 2011

### roam

1. The problem statement, all variables and given/known data

[PLAIN]http://img718.imageshack.us/img718/3508/questionigu.jpg [Broken]

2. Relevant equations

$\epsilon = N \frac{\Delta \Phi_B}{\Delta t} = N \frac{\Delta (BA)}{\Delta t}=NA\frac{\Delta B}{\Delta t}$

3. The attempt at a solution

The correct solution must be 6.54 mV.

So we are told 1000 revolutions per minute so the coil is rotated 1000 times through 180 degrees in 60 seconds. That means the magnetic field changes from 2x10-5 to -2x10-5 in 0.06 seconds since

$\frac{1000}{60}=\frac{1}{x} \implies x = 0.06$

So we have

$\frac{dB}{dt} = \frac{(2\times10^{-5})-(-2\times 10^{-5})}{0.06} = 6.67 \times 10^{-4}$

And the area is 0.25x0.25=0.625 m2, so substituting into the above equation:

$\epsilon = 50 \times (6.67 \times 10^{-4}) \times 0.625 = 2.081 \ mV$

So, what is my mistake? Any help is appreciated.

Last edited by a moderator: May 5, 2017
2. Oct 21, 2011

### SammyS

Staff Emeritus
You calculated the average voltage over the period of 1/2 a revolution.

The voltage will vary sinusoidally (like a sine wave).

Look $\frac{d}{dt}\left(\vec{B}\cdot\vec{A}\right)$.

$\vec{B}\cdot\vec{A}=\left|\vec{B}\right| \,\left|\vec{A}\right|\cos(\theta)$

The angle, θ, changes at a constant rate, which can be determined by the rotational rate of the coil.

BTW: You rounded off 1000/60, too much (and incorrectly) considering it's an intermediate calculation, used to calculate other quantities.

There is no round off error where I said there was one. DUH !

Last edited: Oct 22, 2011
3. Oct 22, 2011

### roam

What do I exactly need to do? I'm a bit confused because I used the time for 1 revolution not 1/2 of a revolution. And I didn't round it off intentionally, my calculator gives 1/16.6666..7=0.06.

4. Oct 22, 2011

### SammyS

Staff Emeritus
(There is no rounding error.)

The fact that you used the time of one revolution rather than 1/2 a revolution means that your answer makes more sense to me now. The average voltage should be about 60% of the peak voltage and your answer was significantly less that 50% of the correct answer.

Convert the rotational velocity (in rev. per minute) to angular velocity (in radians per second).

The angular velocity is dθ/dt .

5. Oct 22, 2011

### roam

I'm not sure how to do that at the moment, but how does that help? Don't we already have the time interval we need?

6. Oct 22, 2011

### SammyS

Staff Emeritus
dΦ/dt, is not constant during even 1/ a rotation. dΦ/dt varies sinusoidally. You need to find the maximum value of dΦ/dt

Look at my previous post. $\left|\vec{A}\right|$ and $\left|\vec{B}\right|$ are constant. All that varies is cos(θ)

What is $\displaystyle \frac{d(\cos(\theta))}{dt}\,?$

7. Oct 22, 2011

### roam

I don't get it. The angle θ between B and the normal to the loop can change with time, so what value do I have to use for it? Do I just have to convert "1000 rev/min" to radians per second?

Last edited: Oct 22, 2011
8. Oct 22, 2011

### SammyS

Staff Emeritus
Yes. That's a great idea. That is the rate at which the angle between B and the normal to the loop is changing assuming the rotational speed is constant.

dθ/dt is in units of radians per second.

9. Oct 22, 2011

### roam

I tried that, didn't work. 1000 RPM is 0.2908 rad/s.

$\epsilon = \frac{d}{dt}(BA cos \theta) = \frac{(2\times10^{-5})\times 0.625 \times cos \ 0.2908}{0.06} = 2.083 \times 10^{-5}$

So how come it doesn't work?

10. Oct 22, 2011

### SammyS

Staff Emeritus
Actually, it does work. You need to take the derivative of cos(θ)

$\displaystyle \frac{d}{dt}\left(BA\cos(\theta)\right)=-AB\sin(\theta)\frac{d\theta}{dt}$

This gives the induced EMF, per loop, as a function of θ, which is (implicitly) a function of time, t. For what value of θ is the induced EMF a maximum?

As mentioned in post #4, "The angular velocity is dθ/dt . " You have found that dθ/dt ≈ 0.2908 rad/s .

Last edited: Oct 22, 2011