Question

Two long solenoids of radii r₁ and r₂ (>r₁) and number of turns per unit length n₁ and n₂ respectively are co-axially wrapped one over the other. The ratio of self-inductance of inner solenoid to their mutual inductance is ______.

Options

• `n_1/n_2`

• `n_2/n_1`

• `(n_1r_1^2)/(n_2r_2^2)`

• `(n_2r_1^2)/(n_1r_2^2)`

Answer 1

Two long solenoids of radii r₁ and r₂ (>r₁) and number of turns per unit length n₁ and n₂ respectively are co-axially wrapped one over the other. The ratio of self-inductance of inner solenoid to their mutual inductance is `bbunderline(n_1/n_2)`.

Explanation:

Given: Inner solenoid:

Radius = r1

Number of turns per unit length = n1

Length = l    ...(assuming both solenoids have the same length)

Outer solenoid:

Radius = r₂ where r₂ > r₁

Number of turns per unit length = n2

The self-inductance of a solenoid is given by:

L = μ₀n₂Al

For the inner solenoid, the cross-sectional area is A₁ = `pir_1^2`

So, L₁ = `mu_0n_1^2pir_1^2l`

Mutual inductance is given by:

M = `mu_0n_1n_2A_"common"l`

Since the inner solenoid is completely enclosed by the outer solenoid, the common cross-sectional area is the area of the inner solenoid, i.e., `A_"common" = pir_1^2`. Thus, the mutual inductance is:

M = `mu_0n_1^2n_2pir_1^2l`

Required ratio:

`L_1/M = (mu_0n_1^2pir_1^2l)/(mu_0n_1n_2pir_1^2l)`

`L_1/M = n_1/n_2`