JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Electromagnetic Induction

There are two long co-axial solenoids of 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:

Answer: (D)

Solution

For the inner solenoid carrying the current, the flux it links with the outer coil spans only the inner cross-section

π r_{1}^{2}

, so the mutual inductance is

M = μ_{0} π r_{1}^{2} n_{1} n_{2} ℓ

. Its own self-inductance is

L = μ_{0} π r_{1}^{2} n_{1}^{2} ℓ

. Dividing, the common factor

μ_{0} π r_{1}^{2} n_{1} ℓ

cancels and

\frac{M}{L} = \frac{n _{2}}{n _{1}}

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option