Author Topic: Electric Power Generation (Read 1596 times) Tweet Share

iamdan08 · « **on:** July 31, 2008, 06:10:22 pm »

Any help with this question would be great

A simple generator consists of a coil of $N=1000$ turns each of radius $10 cm$ , mounted on an axis in a uniform magnetic field of strength $B$ . The coil is rotated at a frequency of $50 Hz$ . Calculate the strength of the magnetic field recquired to produce a voltage of $8.0 kV$

dcc · « **Reply #1 on:** July 31, 2008, 06:48:24 pm »

$\mathcal{E}_{avg} = N \dfrac{\Delta \phi}{T} = N \Delta \phi \cdot f$ since ( $T = \frac{1}{f}$ )

Since $\Delta \phi = B \cdot A = B \cdot 0.01\pi \mbox{ Wb}$ (since $A = \pi r^2$ , assuming that since you referred to these coils having a 'radius' that they are circular)

$\mathcal{E}_{avg} = 8000 V =1000 \cdot B \cdot 0.01 \pi \cdot 50$

$B = \dfrac{8000}{1000 \cdot 0.01 \pi \cdot 50} = \dfrac{16}{\pi}\ T$

(i hope)

iamdan08 · « **Reply #2 on:** July 31, 2008, 06:56:02 pm »

The back of the book says: 0.81 T

dcc · « **Reply #3 on:** July 31, 2008, 07:07:07 pm »

No clue how they got that answer, I'm afraid I cant help any further for lack of further details

iamdan08 · « **Reply #4 on:** July 31, 2008, 07:19:40 pm »

The only other info given is a voltage(kV) - time graph, which if you want to sketch is the same as the equation $y=8sin(100\pi*x$ )

Mao · « **Reply #5 on:** July 31, 2008, 08:48:14 pm »

right. This means that the peak is 8kV

hence the formula $\mathcal{E}_{max}=N(BA)(2\pi f)$ will be useful to you

alternatively (just for fun)

$V(t)=8000\sin(100\pi t)$

$\Phi(t)=\frac{1}{N}\int -V(t)\; dt = \frac{2}{25\pi}\cos(100\pi t)$

$\Phi_{max}=\frac{2}{25\pi} wb=BA$

$\implies B=\frac{2}{25\pi}\cdot \frac{1}{A}=\frac{2}{25\pi}\cdot \frac{100}{\pi}=\frac{8}{\pi^2}T$

$\implies B\approx 0.81T$

iamdan08 · « **Reply #6 on:** August 01, 2008, 09:11:33 am »

Cheers mao. Nice method!

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