Question

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

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

Answer: (A)

The 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 $\times$ area of cross section where $B\ne 0$
= $\frac{{\mu }_0{n}_2^2{i}_2^2}{2}\pi \left[r_2^2-r_1^2\right]$
$=\frac{{\mu }_0{n}_1^2{i}_1^2}{2}\left[\pi r_2^2-r_1^2\right]$ , since $n_1{i}_1=n_2{i}_2$