JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Electromagnetic Induction

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The total number of turns and cross-section area in solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:

(A) $1/ L$ ✓ Correct
(B)L
(C) $1/ L^{2}$
(D) $L^{2}$

Answer: (A)

Solution

The inductance of a solenoid is

L = μ_{0} n^{2} A ℓ

, with

n

the turns per unit length. Holding the total turns

N

and the area

A

fixed,

n = \frac{N}{ℓ}

, so

L = μ_{0} \frac{N ^{2}}{ℓ ^{2}} A ℓ = μ_{0} \frac{N ^{2} A}{ℓ}

. With

N

and

A

constant this leaves

L \propto \frac{1}{ℓ}

, so stretching the solenoid lowers its inductance.

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option

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