Question

Calculate the mutual induction coefficients of two concentric coils having number of turns 100 and 200, radii 5 cm, 10 cm and carrying currents 2 A and 3A respectively

Answer 1

Explanation:

Calculate the mutual induction coefficients of two concentric coils having number of turns 100 and 200, radii 5 cm, 10 cm and carrying currents 2 A and 3A respectively

Answer 2

Answer:

The mutual inductance of the given coil is M = 3.14 x 10^-3 H

Explanation:

It is given that the :

• the Number of turns = 100 and 200 respectively

• the radius of the coils = 5 cm and 10 cm respectively

• given Currents = 2 A and 3 A respectively

we know that :

Mutual Inductance of the coil = M.i = N1 . Ф

where, Ф = Magnetic field persisting x Area of the substance

therefore, Magnetic field of the coil = N2μoi / 2r2

and the respective area = π(r1)^2

Therefore, Mutual inductance of the coil = N1 x (N2μoi / 2r2) x  π(r1)^2

Cancelling the value of currents from both sides, we get

M = N1N2μo π(r1)^2 / 2r2

and we know that μo = 4π x 10^-7

Therefore, M = 100 x 200 x 4π x 10^-7 x 0.05^2 / 2 x 0.1^2

which gives us that M = 3.14 x 10^-3 H

therefore the mutual inductance of the given coil is M = 3.14 x 10^-3 H

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