Question

A 0.270 m radius, 510-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.

Answer 1

Answer:

0.35701 T

Explanation:

$B_i$ = Initial magnetic field

$B_f$ = Final magnetic field

$\phi$ = Magnetic flux

t = Time taken = 4.17 ms

N = Number of turns = 510

$\epsilon$ = Induced emf = 10000 V

r = Radius = 0.27 m

A = Area = $\pi r^2$

Induced emf is given by

$\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T$

The magnetic field strength needed is 0.35701 T