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.

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:

[tex]emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}[/tex] (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):

[tex]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)}\\\\[/tex]

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

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

The strength of the magnetic field is 0.37T