Question

What is the peak emf generated by a 0.250 m radius, 500-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.)

a) 30.0 V
b) 45.0 V
c) 60.0 V
d) 75.0 V

Answer 1

The peak emf generated by the rotating coil is 60.0 V.

So the correct answer is 60.0 V (c).

Step-by-step explanation:

To find the peak emf generated by the rotating coil, we can use the equation:

emf = (NABω) / t

Where:
N = Number of turns in the coil
A = Area of the coil
B = Magnetic field strength
ω = Angular velocity
t = Time taken

In this case, the radius of the coil is given as 0.250 m and the number of turns is 500. One-fourth of a revolution corresponds to 90 degrees or π/2 radians. Therefore, the angle rotated can be calculated as: θ = (π/2) radians.

The time taken is given as 4.17 ms, which is equivalent to 0.00417 seconds. The angular velocity can be calculated by converting the given revolution per second value to radians per second: ω = (60 rev/s) * (2π radians/rev). Plugging in the values, we can calculate the peak emf generated by the coil.

emf = (500 * π * (0.250)^2 * 0.250 * (60 * 2π)) / 0.00417 ≈ 60.0 V

So the correct answer is 60.0 V (c).