Question

A 550-turn solenoid, 28cm long, has a diameter of 2.9cm . A 16-turn coil is wound tightly around the center of the solenoid...If the current in the solenoid increases uniformly from 0 to 4.2A in 0.50s , what will be the induced emf in the short coil during this time? pleased help and explain..:)

Answer 1

Given that number of turns is N = 550

Length of the solenoid is L = 28 cm

diameter is d = 2.9 cm

radius of the coil is r = 1.45 cm

Area of the coil is A = ?r^2 = ? ( 1.45*10^-2m)^2 = 6.6*10^-4 m^2

magnetic field is B = ?_0 n I

    n = N_1 / L

       = 550/ 28 *10^-2 m

       = 1964.285 turns / m

Hence, the magnetic field is, B = 4?*10^-7 * 1964.285* 4.2 A = 1.036*10^-2 T

the induced emf   ? = N d ? / dt = 16 * 1.036*10^-2 T * 6.6*10^-4 m^2 / 0.50 s = 2.19*10^-4 V