Let us consider two solenoids A and B , made from same magnetic material of relative permeability μ_r and equal area of cross-section. Length of A is twice that of B and the number of turns per unit length in A is half

Correct answer: A. 1: 2 L=μ_0 μ_r × n × A × N L=μ_0 μ_r n × A × (N)/(l) × l L=μ_0 μ_r × n^2 × A × I L n^2 I (L_A)/(L_B)=(n_A^2)/(n_B^2) × (I_A)/(I_B) (L_A)/(L_B)=(1)/(4) × 2=(1)/(2)

NEET 2024

Exam: NEET
Year: 2024
Subject: Physics
Chapter: Electromagnetic Induction

NEET 2024 · Physics · Electromagnetic Induction

Let us consider two solenoids A and B , made from same magnetic...

Let us consider two solenoids

A

and

B

, made from same magnetic material of relative permeability

\mu_r

and equal area of cross-section. Length of

A

is twice that of

B

and the number of turns per unit length in

A

is half that of

B

. The ratio of self inductances of the two solenoids,

L_A: L_B

Options

A $1: 2$
B $2: 1$
C $8: 1$
D $1: 8$

Correct answer

A. $1: 2$

Step-by-step solution

L=\mu_0 \mu_r \times n \times A \times N

L=\mu_0 \mu_r n \times A \times \frac{N}{l} \times l

L=\mu_0 \mu_r \times n^2 \times A \times I \quad \Rightarrow L \propto n^2 I

\Rightarrow \frac{L_A}{L_B}=\frac{n_A^2}{n_B^2} \times \frac{I_A}{I_B}

\Rightarrow \frac{L_A}{L_B}=\frac{1}{4} \times 2=\frac{1}{2}

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