# Coil Magnetic Field Question

1. Apr 9, 2014

### patrickmoloney

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

A planar 50-turn coil of area $0.20m^2$ is rotated in a uniform $0.15$ Tesla magnetic field by a motor at a constant angular velocity $\omega = 315 s^-1$ The axis of rotation is in the plane of the coil and is perpendicular to the magnetic field. At time $t = 0$, the magnetic field is perpendicular to the plane of the coil.

B.
(i) Plot the total magnetic flux through the coil as a function of time. [3 marks]
(ii) Find the maximum emf generated from the coil. [3 marks]
(iii) What is value of the root-mean-square emf? [3 marks]

C.
If the rotating coil in part (b) forms part of an electrical circuit with total resistance $30 \Omega$(including the coil),

(i) What is the average power generated? [3 marks]
(ii) Plot the instantaneous power as a function of time. [3 marks]

2. Relevant equations

$\epsilon = -N \frac{d\Phi_B}{dt}$

$P = I^2R$

$\Phi_B = BAcos(\omega t)$

3. The attempt at a solution

(i) I need help with this part, I know Flux goes on the y-axis and Time on the x-axis. I'm not sure how to go about plotting the points. I think $\omega$ depends on the amplitude of the flux.

(ii) Maximum emf $(\epsilon)$ when $\theta = 90°$ i.e when $\omega t = 90°$

$\Phi = NBA(cos\theta)$

$\epsilon = -N \frac{d\Phi}{dt} = -NBA \frac{d}{dt}(cos \theta) = -NBA \frac{d}{dt}(cos (\omega t))$

$\epsilon = -NBA(-\omega sin(\omega t))$ , where $\omega t = 90°$

$\epsilon = NBA\omega = (50)(0.15)(0.2)(314) = 471 V$

(iii) $\epsilon_{rms} = \sqrt{\frac{NBA\omega}{2}}$

$= \sqrt{\frac{(50)(0.15)(0.2)(314)}{2}} = 15.35 V$

C.

(i) $P = I^2R$ Ohm's Law states $E=IR \Rightarrow I^2 = \frac{E^2}{R^2}$

Now, $P = \frac{E^2}{R} = \frac{(471)^2}{30} = 7394.7 W$

(ii) As you can see I really struggle with graphs, how do I go about plotting the instantaneous powers as a function of time?

2. Apr 9, 2014

### BvU

Hello Patrick.
Your "I think ω depends on the amplitude of the flux" surprises me. In the problem statement it says that ω = 315 /s, so why should it not be that value ?
And in (ii) you use $\Phi$ correctly, as far as I can see.
Then again, in (iii) you surprise me: no suspicion if the rms of an alternating voltage with an amplitude of 471 V comes out so extremely low ?

Next, C (i): Is that the average power or the amplitude of the instantaneous power ?

3. Apr 9, 2014

### patrickmoloney

Sorry I meant, I think the flux is dependent on the amplitude of omega. In B (iii) I thought that was the formula for the root-mean square of emf. I obtained it online, since I could not find it in my text book. C (i) $P_{avg} = \frac{V_{rms}^2}{R}$ Where $V_{rms} = \frac{V}{\sqrt{2}}$ is this the average power?

4. Apr 10, 2014

### BvU

Still don't understand what is blocking you: you write $\Phi_B = BAcos(\omega t)$, they tell you B, they tell you A, they tell you $\omega$ and all you have to do is plot a cosine and put the right values at the tick marks. The amplitude of the cosine comes in, but not the "amplitude" of $\omega$.

B(iii) formula is fine (check what it means); my comment was more that you should be suspicious if you get 15 when dividing 471 by $\sqrt{2}$

C (i): yes.

rms is a simple thing for a sine wave: squaring gives sin2 which averages out to 1/2. taking the root gives you ${1\over 2}\sqrt{2}$.

For e.g. average power the squaring indeed comes from $P = V \cdot I = V^2/R = I^2 R$ for a simple resistance (where there is no phase difference between V and I).

With which I am opening a can of worms, because here we have a coil, that not only has a resistance (a part of the 30 $\Omega$) but also a self-inductance that is introducing a phase difference between V and I. If you haven't been introduced to that stuff already, you might get away with ignoring it. After all, they tell you that the circuit has a total resistance of 30 $\Omega$.

 Disregard last paragraph. L ≠ 0 clearly not intended. Any expert's comment on this issue ?

Last edited: Apr 10, 2014