Question

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, LA : LB is

• A. 1 : 2
• B. 2 : 1
• C. 8 : 1
• D. 1 : 8

Answer 1

The Correct Option is A

1. Self-inductance of a solenoid 

For a long solenoid, self-inductance $L$ is:

$$ L = \mu_0 n^2 A l = \mu_0 n^2 \times \text{volume} $$

So, $L \propto n^2 l$, where:

• $n$ = number of turns per unit length
• $l$ = length of solenoid
• $A$ = cross-sectional area (assumed same for both)

2. Given relations

For solenoids A and B:

$$ n_A = \frac{1}{2} n_B \quad \text{and} \quad l_A = 2 l_B $$

Also given: $L_A = 2L_B$. Let’s verify using $L \propto n^2 l$.

3. Take the ratio

$$ L_A \propto n_A^2 l_A \quad , \quad L_B \propto n_B^2 l_B $$

$$ \frac{L_A}{L_B} = \frac{n_A^2 l_A}{n_B^2 l_B} $$

Substitute $n_A = n_B/2$ and $l_A = 2l_B$:

$$ \frac{L_A}{L_B} = \frac{(n_B/2)^2 \cdot (2l_B)}{n_B^2 \cdot l_B} $$

$$ \frac{L_A}{L_B} = \frac{\frac{n_B^2}{4} \cdot 2l_B}{n_B^2 l_B} = \frac{2}{4} = \frac{1}{2} $$

Thus, $L_A : L_B = 1 : 2$

Note: This contradicts the given $L_A = 2L_B$. Based on $L \propto n^2 l$, the correct ratio should be $1:2$.