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

Answer: 3.141 X $10^-3$ H

Explanation:

Mutual Inductance= M.i = $N_1$ . Ф

, where $N_1$  = Number of turns of inner coil & Ф = flux

Now,

Ф = Magnetic field X Area

Magnectic field due to outer coil will be  $\frac{N_2u_0i}{2r_2}$ &

Area = $\pi (r_1)^2$

Substituting these values in the Mutual inductance formula, we get

∴ M.i = $N_1$ . ( $\frac{N_2u_0i}{2r_2}$ ).$\pi (r_1)^2$

Cancelling i(current) from both the sides of the above equation, we have

M = $\frac{N_1N_2u_0\pi(r_1)^2 }{2r_2}$

$u_0$ = 4$\pi$ X $10^{-7}$

M = $\frac{100.200.4\pi X(10^-7).0.05^2 }{2 X0.1^2}$

M = 3.141 X $10^-3$ H

Answer 2

Thus the mutual inductance is M = 3.14 x 10^-3 H

Explanation:

Given data:

• Number of turns = 100 and 200
• Radii of coils = 5 cm and 10 cm
• Currents = 2 A and 3 A

Solution:

Mutual Inductance= M.i = N1 . Ф

Ф = Magnetic field x Area

Magnetic field = N2μoi / 2r2

Area = π(r1)^2

Put these values in the formula:

Mutual inductance = N1 x (N2μoi / 2r2) x  π(r1)^2

Current gets cancelled on both sides.

M = N1N2μo π(r1)^2 / 2r2

μo = 4π x 10^-7

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

M = 3.14 x 10^-3 H

Thus the mutual inductance is M = 3.14 x 10^-3 H