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

0.35701 T

Step-by-step 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(d\phi)/(dt)\\\Rightarrow \epsilon=-N(B_fAcos90-B_iAcos0)/(dt)\\\Rightarrow \epsilon=N(B_iA)/(dt)\\\Rightarrow B_i=(\epsilon dt)/(NA)\\\Rightarrow B_i=(10000 * 4.17* 10^(-3))/(510* \pi 0.27^2)\\\Rightarrow B_i=0.35701\ T`

The magnetic field strength needed is 0.35701 T