Let us consider two solenoids $A$ and $B$, made from same magnetic material of relative permeability

Question

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, $L_A: L_B$ is

Professor PC Thomas Classes · Answer

(a) : Self inductance of solenoid, $$ L=\frac{\mu_o N^2 A}{l}=\mu_r \mu_o n^2 l A $$ Where, $n$ is number of turns per unit length $A=$ Area and, $l=$ length Given: $l_A=2 l_B, n_A=\frac{1}{2} n_B$ $$ 	herefore \quad \frac{L_A}{L_B}=\frac{\mu_r \mu_o l_A A_A n_A^2}{\mu_r \mu_o l_B A_B n_B^2}=2 l_B \cdot \frac{n_B^2}{4 n_B^2 l_B}=\frac{1}{2} \quad\left(\because A_A=A_Bight) $$