Question

A solenoid with A=4.0x10^-4m^2 and length of 25cm is constructed with 300 turns. Calculate the inductance and the self inducted emf if a current change of 50 A/s flows.

Answer 1

here ,

di/dt = 50 A/s

A = 4 *10^-4 m^2

L = 25 cm = 0.25 m

N = 300

inductance = u0 * N^2 * A/L

inductance = 4pi *10^-7 * 300^2 * 4 * 10^-4/.25

inductance = 1.81 *10^-4 H

the inductance is 1.81 *10^-4 H

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self induced emf = inductance * di/dt

self induced emf = 1.81 *10^-4 * 50

self induced emf = 0.00905 V

the self induced emf is 0.00905 V