Question

A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25 T field, starting with the plane of the coil parallel to the field. What is the peak emf?

(a) 150 V
(b) 300 V
(c) 450 V
(d) 600 V

Answer 1

The peak emf generated by the coil can be calculated using Faraday's Law, and in this case it is approximately 147.26 V, which corresponds to answer choice (a) 150 V.

Step-by-step explanation:

The question involves calculating the peak electromagnetic force (emf) generated by a rotating coil in a magnetic field. Using Faraday's Law of Induction, the peak emf (ε) can be determined with the formula ε = NABωsinωt, where N is the number of turns,

A is the area of the coil, B is the magnetic field strength, ω is the angular velocity, and t is the time. In this scenario, the coil has 75 turns (N=75), a diameter of 0.10 m (thus radius r=0.10/2 meters and area A = πr^2), is in a 1.25 T magnetic field (B=1.25 T), and has an angular velocity of 8.00 rad/s (ω=8.00 rad/s). The peak emf is when sinωt = 1, so we ignore sinωt when calculating peak emf. The calculated peak emf is:

Peak emf = NABω = 75 * π * (0.05 m)^2 * 1.25 T * 8.00 rad/s = 147.26 V, which is closest to choice (a) 150 V, making it the correct option.