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Question

Physics Question on Inductance

There are two long co-axial solenoids of same length ll. the inner and outer coils have radii r1r_1 and r2r_2 and number of turns per unit length n1n_1 and n2n_2 respectively. The rate of mutual inductance to the self-inductance of the inner-coil is :

A

n2n1.r22r12\frac{n_{2}}{n_{1}}. \frac{r^{2}_{2}}{r_{1}^{2}}

B

n2n1.r1r2\frac{n_{2}}{n_{1}}. \frac{r_{1}}{r_{2}}

C

n1n2\frac{n_1}{n_2}

D

n2n1\frac{n_2}{n_1}

Answer

n2n1\frac{n_2}{n_1}

Explanation

Solution

M=μ0n1n2πr12M = \mu_{0} n_{1}n_{2} \pi r^{2}_{1} L=μ0n12πr12L =\mu_{0} n_{1}^{2} \pi r^{2}_{1} ML=n2n1\Rightarrow \frac{M}{L} = \frac{n_{2}}{n_{1}}