Two solenoids A and B spaced close to each other and sharing

Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 450 and 740 turns, respectively. A current of 4.00 A in solenoid A produces an average flux of 300 µWb through each turn of A and a flux of 90.0 µWb through each turn of B. (a) Calculate the mutual inductance of the two solenoids. mH (b) What is the inductance of A? mH (c) What is the magnitude of the emf that is induced in B when the current in A changes at the rate of 0.500 A/s? mV

Solution

Here,

I = 4 A

N1 = 450

N2 = 740

a) let the mutual inductance of solenoids is M

flux = M* I

740 * 90 *10^-6 = M * 4

M = 0.0163 H

the mutual inducance is 0.0163 H

b)

let the inductance of A is L

flux = 450 * L * I

450 * 300 *10^-6 = L * 4

L = 0.03375 H

c)

let the voltage is V

VA = M * di/dt

VA = 0.0163 * 0.50

VA = 8.15 *10^-3 V

the emf induced in A is 8.15 *10^-3 V