Question

A 1200 - turn solenoid is coaxial with a single-turn conducting ring as shown in the figure. The length of the solenoid is 20.0 cm, and the current applied to it varies in time according to the relation I(t) = I_0(1 - e^-t/tau), where I_0 = 3.00 A and tau = 2.00 s. The diameter of the solenoid is 4.00 cm, whereas that of the conducting ring is 8.00 cm. If the resistance of the ring is 0.50 Ohm, compute the magnitude and direction induced current in the ring at 3.00 s.

Answer 1

I(t) = I_0 (1 - e^- t/z) For solenoid B = mu_0^nI = mu_0n I_0 (1 - e^- t/z) = mu_0 n 3 (1 - e^- t/z) (phi_B) = integral Bdx = mu_0 n 3 (1 - e^- t/z) integral dA = mu_0 n 3 (1 - e^- t/z) pi R^2 From Faraday�s law emf = - N d phi 3/ dt = - N mu_0 n 3 (1 - e^- t/z) pi R^2 (- 1/2) = mu pi times 10^- 7 times (1) times 1200/0.20 times 3 (1 - r^- 0.5t) pi (0.04)^2 times 0.5 emf = 5.684892 times 10^- 5 (1 - e^- 0.5t) From ohm�;s law, I = emf/R = 5.684892 times 1^- 5 (1 - e^- 0.5 (3))0/.5 I = 8.8328 times 10^- 5 A By Lent�s law, direction st, current