Question

An electric generator has a coil consisting of 50 turns with a cross-sectional area of 0.4 m². The generator turns at a rate of 40 rad/s in a magnetic field of 0.1 T I magnitude. What is the peak emf produced by the generator?

a) 0 V

b) 40 V

c) 80 V

d) 120 V

e) 160 V

Answer 1

The peak emf produced by the electric generator is calculated using the formula E = NABω, which yields a result of 80 V. Thus, the correct answer is (c) 80 V.

Step-by-step explanation:

The student's question pertains to the calculation of peak emf produced by an electric generator. According to Faraday's law of electromagnetic induction, the peak emf (E) induced in a coil is given by the equation E = NABωsin(ωt), where N is the number of turns in the coil, A is the area of the coil, B is the magnetic field strength, and ω is the angular velocity. Given that sin(ωt) reaches its maximum value of 1 at the peak emf, we can calculate the peak emf using the given values: N = 50, A = 0.4 m², B = 0.1 T, and ω = 40 rad/s.

The peak emf then is E = NABω = 50 × 0.4 m² × 0.1 T × 40 rad/s = 80 V.

Therefore, the correct answer is (c) 80 V.