There are two long co-axial solenoids of same length l . The inner and outer coils have radii r₁ and r₂ and number of turns per unit length n₁ and n₂respectively. The ratio of mutual inductance to the self inductance of the inner – coil is

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$\frac{n_{2}}{n_{1}} . \frac{r_{2}^{2}}{r_{1}^{2}}$

$\frac{n_{1}}{n_{2}}$

$\frac{n_{2}}{n_{1}} . \frac{r_{1}}{r_{2}}$

$\frac{n_{2}}{n_{1}}$

answer is A.

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Detailed Solution

The mutual inductance of the inner coil
$M = μ_{0} n_{1} n_{2} {πr}_{1}^{2} l$ ...............(1)
Self inductance (L) of the inner coil is $L = μ_{0} n_{1}^{2} {πr}_{1}^{2} l$ ………….(2)
Using (i) and (ii) ,
$\frac{M}{L} = \frac{n_{2}}{n_{1}}$