Question

A 0.260 m radius, 525 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:

B = 0.37T

Explanation:

In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:

$emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}$       (1)

emf: induced voltage in the solenoid = 10,000V

N: turns of the solenoid = 525

ФB: magnetic flux

B: magnitude of the magnetic field = ?

A: cross-sectional area of the solenoid = π*r^2

r: radius of the cross-sectional area = 0.260m

Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s

First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90°  (quarter of a revolution)

In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):

$emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}$

Finally, you replace the values of the parameters to calculate B:

$B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T$

The strength of the magnetic field is 0.37T