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

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Solution

Q.
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 that of $B .$ The ratio of self inductances of the two solenoids, $L_{A} : L_{B}$ is [2024]

1 1 : 2

2 2 : 1

3 8 : 1

4 1 : 8

Ans.
(1)

Self inductance of solenoid,

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

Where, $n$ is number of turns per unit length,

$A =$ Area and, $l =$ length

Given: $l_{A} = 2 l_{B}, n_{A} = \frac{1}{2} n_{B}$

$∴$ $\frac{L_{A}}{L_{B}} = \frac{μ_{r} μ_{0} l_{A} A_{A} n_{A}^{2}}{μ_{r} μ_{0} l_{B} A_{B} n_{B}^{2}} = 2 l_{B} \cdot \frac{n_{B}^{2}}{4 n_{B}^{2} l_{B}} = \frac{1}{2}$ $(∵ A_{A} = A_{B})$

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