Question

Two solenoids of equal number of turns having their length and the radii in the same ratio $1:2$. The ratio of their self-inductance will be

• A. $1:2$
• B. $2:1$
• C. $1:1$
• D. $1:4$

Answer 1

The correct option is A $1:2$

Self inductance of a solenoid,
$L=\frac{{\mu }_0{N}^{2}A}{l}=\frac{{\mu }_0{N}^2\pi r^2}{l}$
where $l$ is the length of the solenoid, N is the total number of turns of the solenoid and A is the area of cross-section of the solenoid.
$\therefore \frac{L_1}{L_2}={\left(\frac{N_1}{N_2}\right)}^2{\left(\frac{r_1}{r_2}\right)}^2\left(\frac{l_2}{l_1}\right)$
Here, $N_1=N_2,\frac{l_1}{l_2}=\frac{1}{2},\frac{r_1}{r_2}=\frac{1}{2}$
$\therefore \frac{L_1}{L_2}={\left(\frac{1}{2}\right)}^2\left(\frac{2}{1}\right)=\frac{1}{2}$