A solenoidal coil with 30 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 24.0 cm long and has a diameter of 2.50 cm. At a certain time, the current in the inner solenoid is 0.140 A and is increasing at a rate of 1700 A/s. For this time, calculate the average magnetic flux through each turn of the inner solenoid. For this time, calculate the mutual inductance of the two solenoids; For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Explanation / Answer

flux of inner solenoid phi = AB cos theta

B due to Solenoid = uo N i/L

FLux = 3.14 * 0.0125 * 0.0125 * 4*3.14*10^-7 * 300 *0.14/0.24

flux = 1.07*10^-7 Wb

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mutual inductance M = pi uo Ns Np rs^2/2Rp

where Ns = no. or turns in secondary coil

Np = mo. of turns per unit length in primary coil

rp is the raidu of primary coil

rs = raidus of seconday coil

M = 4*3.14*10^-7 * 300 * 30 * 3.14 * 0.0125^2/(2 * 0.24)

M = 11.55 uH or 1.15 *10^-5 H

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incuded emf e = L di/dt

e = 1.15 *10^-5 * 1700

e = 0.01955 Volts

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