Question

Exercise 30.12 - Enhanced - with Feedback Along, straight solenoid has 700 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is 3.25 x 10-3 Wb. Part A What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.50 mV? Express your answer in milliamperes per second. ANSWER: die mA/s

Answer 1

(A) Inducatance of solenoid, L = Flux/Current

= (3.25*10^-3/2.90) Henry

= 1.120 H

Now, induced emf = NL(dI/dt)

Or, 6.50*10^-3 = 700*1.120*(dI/dt)

Or, dI/dt = (6.50*10^-3/784)

Thus, dI/dt = 0.00829*10^-3 A/s = 0.00829 mA/s