An AC generator consists of 6 turns of wire. Each turn has an area of 0.040 m2. The loop rotates in a uniform field (B = 0.20 T) at a constant frequency of 50 Hz. What is the maximum induced emf?

The maximum induced emf is 15.08 V

Explanation:

Given;

number of turns of the generator, N = 6 turns

area of the loop, A = 0.04 m

magnetic field of the loop, B = 0.2 T

frequency of loop, f = 50 Hz

The  maximum induced emf is given by;

E = NBAω

Where;

ω is the angular speed = 2πf

E = NBA(2πf)

E = 6 x 0.2 x 0.04 x (2 x 3.142 x 50)

E = 15.08 V

Therefore, the maximum induced emf is 15.08 V

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.