Question

Two coaxial long solenoids of equal lengths have current, $i_1$, $i_2$, number of turns per unit length $n_1,n_2$ and radius $r_1,r_2$ respectively. If $n_1{i}_1=n_2{i}_2$ and the two solenoids carry current in opposite sense, the magnetic energy stored per until length is $[r_2>r_1]$

• A. $\frac{{\mu }_0}{2}{n}_1^2{i}_1^2\pi (r_2^2-r_1^2)$
• B. ${\mu }_0{n}_1^2{i}_1^2\pi (r_2^2-r_1^2)$
• C. $\frac{{\mu }_0}{2}{n}_1^2{i}_1^2\pi r_1^2$
• D. $\frac{{\mu }_0}{2}{n}_2^2{i}_2^2\pi r_2^2$

Answer 1

The correct option is A $\frac{{\mu }_0}{2}{n}_1^2{i}_1^2\pi (r_2^2-r_1^2)$

$\text{Solution:-}$

Magnetic field is non zero only in the region between the two solenoids where,

$B={\mu }_0{n}_2{i}_2$

$\therefore$ energy stored per unit volume =$\frac{B_2}{2{\mu }_0}=\frac{{\mu }_0{n}_2^2{i}_2^2}{2}$

The energy per unit length=energy per unit volume x area of cross section where,

$B\ne 0$

$=\frac{{\mu }_0{n}_2^2{i}_2^2}{2}\pi \left[r_2^2-r_1^2\right]=\frac{{\mu }_0{n}_1^2{i}_1^2}{2}\pi \left[r_2^2-r_1^2\right]$ , since $n_i{i}_1-n_2{i}_2$

$\text{Hence the correct option is A}$