Question

There are two long coaxial solenoids of same length l. The inner and outer coils have radii r₁ and r₂ and number of turns per unit length n₁ and n₂, respectively. The ratio of mutual inductance to the self-inductance of the inner coil is
1. \(\frac{{{n_2}}}{{{n_1}}}\frac{{{r_1}}}{{{r_2}}}\)
2. \(\frac{{{n_2}}}{{{n_1}}}\frac{{r_2^2}}{{r_1^2}}\)
3. \(\frac{{{n_2}}}{{{n_1}}}\)
4. \(\frac{{{n_1}}}{{{n_2}}}\)

Answer 1

Correct Answer - Option 3 : \(\frac{{{n_2}}}{{{n_1}}}\)

Concept:

Mutual Inductance:

Mutual Inductance is the interaction of one coils magnetic field on another coil as it induces a voltage in the adjacent coil.

Given,

Length of coaxial solenoids = l

Radius of the inner coil = r₁

Radius of the outer coil = r₂

Number of turns of the inner coil = n₁

Number of turns of the outer coil = n₂

Mutual inductance for a coaxial solenoid is

\(M = {\mu _0}{n_1}{n_2}\pi r_1^2l\;\)

(for internal coil of radius r₁ )

Self- inductance \(L = {\mu _0}n_1^2\pi r_1^2l\) 

Calculation:

\(\therefore \frac{M}{L} = \frac{{{\mu _0}{n_1}{n_2}\pi r_1^2l}}{{{\mu _0}n_1^2\pi r_1^2l}}\) 

\(\frac{M}{L} = \frac{{{n_1}{n_2}}}{{n_1^2}}\) 

\(\frac{M}{L} = \frac{{{n_2}}}{{{n_1}}}\) 

Therefore, the ratio of mutual inductance to the self-inductance of the inner coil is \(\frac{{{n_2}}}{{{n_1}}}\).