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(D) The self-inductance of the inner solenoid is given by $L_1 = \mu_0 n_1^2 A_1 l = \mu_0 n_1^2 (\pi r_1^2) l$.The mutual inductance of the two solen

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$\frac{n_2}{n_1}$

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(D) The self-inductance of the inner solenoid is given by $L_1 = \mu_0 n_1^2 A_1 l = \mu_0 n_1^2 (\pi r_1^2) l$.The mutual inductance of the two solenoids is given by $M = \mu_0 n_1 n_2 A_1 l = \mu_0 n_1 n_2 (\pi r_1^2) l$,where $A_1$ is the cross-sectional area of the inner solenoid.The ratio of mutual inductance to the self-inductance of the inner coil is $\frac{M}{L_1} = \frac{\mu_0 n_1 n_2 \pi r_1^2 l}{\mu_0 n_1^2 \pi r_1^2 l}$.Simplifying this expression,we get $\frac{M}{L_1} = \frac{n_2}{n_1}$.

There are two long coaxial solenoids of the same length $l$. The inner and outer coils have radii $r_1$ and $r_2$ and number of turns per unit length $n_1$ and $n_2$ respectively. The ratio of mutual inductance to the self-inductance of the inner coil is

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