Question

You decide to build a small generator by rotating a coiled wire inside a static magnetic field of 0.30 T. You construct the apparatus by coiling wire into 5 loops of radius 0.18 m. If the coils rotate at 2.0 revolutions per second and are connected to a device with 120Ω resistance, calculate the average power Paverage supplied to that device.

Answer 1

Answer:0.0153 W

Explanation:

Given

Magnetic Field B=0.3 T

N=5

radius R=0.18 m

$N_0=2 rps$

$\omega =2\pi N_0=12.56 rad/s$

flux associated with loop is

$\phi _B=NBA\cos \theta$

$\phi _B=NBA\cos \omega t$

and Induced emf is

$e=\frac{\mathrm{d} \phi _B}{\mathrm{d} t}$

$e=NBA\omega \sin \omega t$

and let Q be the energy dissipated then

$Q=\int Pdt$

$Q=\int_{0}^{T}\frac{E^2}{R}dt$

$Q=\frac{(NBA\omega )^2}{R}\int_{0}^{T}\frac{1-\cos 2\omega t}{2}dt$

$Q=\frac{(NBA\omega )^2}{R}\times \frac{T}{2}$

Thus $\frac{Q}{T}=P_{avg}$

$P_{avg}= \frac{(NBA\omega )^2}{2R}$

$P_{avg}=\frac{(5\times 0.3\times 0.101/8\times 12.56)^2}{2\times 120}$

$P_{avg}=0.0153 W$