Question

A long, straight solenoid has 860 turns. When the current in the solenoid is 2.00 A , the average flux through each turn of the solenoid is 3.25×10−3Wb.

Part A

What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.00 mV ?

Express your answer to three significant figures and include the appropriate units.

Answer 1

flux = B. A

B = u0 n I

3.25 x 10^-3 = (4pi x 10^-7 x n x 2) (A )

n A = 1293.18

induced emf = d ( N B A)/dt

e = d ( B A ) / dt

6 x 10^-3 = d (u0 n I A )/dt = u0 I A dI/dt

dI/dt = (6 x 10^-3) / (1293.18 x 4 x pi x 10^-7)

= 3.69 A/s