Question

A long, thin solenoid has 870 turns per meter and radius 3.00cm . The current in the solenoid is increasing at a uniform rate of 65.0A/s .

A)What is the magnitude of the induced electric field at a point 0.550cm from the axis of the solenoid?

B)What is the magnitude of the induced electric field at a point 1.00cm from the axis of the solenoid?

Answer 1

a.

The magnetic field due to solenoid is,

           B = u_onI

Differentiate the above equation on both sides,

          dB/dt = u_on dI/dt

The expression for the electric field inside the solenoid is,

       E = r/2 dB/dt

           = (r/2)u_on dI/dt

           = (0.550x10^-2 m /2)(4(pi)x10^-7 T.m/A)(870)(65.0 A/s)

           = 1.95x10^-4 N/C

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b.

The expression for the electric field inside the solenoid is,

       E = r/2 dB/dt

           = (r/2)u_on dI/dt

           = (1.00x10^-2 m /2)(4(pi)x10^-7 T.m/A)(870)(65.0 A/s)

           = 3.55x10^-4 N/C