Question

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 670 turns and solenoid 2 has 450 turns. When the current in solenoid 1 is 6.55 A , the average flux through each turn of solenoid 2 is 4.00×10⁻² Wb.

A-)What is the mutual inductance of the pair of solenoids?

B-)When the current in solenoid 2 is 2.55 A , what is the average flux through each turn of solenoid 1?

Answer 1

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(a)

MMF of solenoid 1 = 670 x 6.55 = 4388.5 A-turns
So, reluctance of the magnetic circuit is:
R = 4388.5/0.04 = 109712.5 A-turn/Wb
Then, supposing a coupling factor of 1, we have the mutual inductance:
M = N1N2/R = (670X450)/109712.5 = 2.74809 H
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(b)

MMF of solenoid 2 = 2.55 x 450 = 1147.5 A-turns
So average flux through each turn of solenoid 1 = 1147.5 /109712.5 = 0.0104592 Wb

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