A 25cmdiameter solenoid is wrapped with 1200 turns per meter

A 2.5-cm-diameter solenoid is wrapped with 1200 turns per meter. 0.60 cm from the axis, the strength of an induced electriefield is 5.2 Times 10^-4 V/m. What is the rate dI/dt with which the current through the solenoid is changing? Express your answer using two significant figures.

Solution

n = 1200

d = 2.5 cm = 0.025 m

Magnetic field inside a solenoid is given by :

B = u*n*I

So, rate of change of magnetic flux:

dB/dt = u*n*dI/dt = induced emf = V

So, Now, V = E*x

So, u*n*dI/dt = E*x

So, dI/dt = E*x/(u*n) = 5.2*10^-4*0.006/(1.26*10^-6*1200)

So, dI/dt = 2.06*10^-3 A/s