Question

A long solenoid with $40$ turns per cm carries a current of $1A$. The magnetic energy stored per unit volume is ........ $J/m^3$.

• A. $3.2\pi$
• B. $32\pi$
• C. $1.6\pi$
• D. $6.4\pi$

Answer 1

Answer: (A)

$n=40$ turns $c{m}^{-1}$
$=40\times 10^2$ turns $m^{-1}$
$I=1.00A$
The magnetic energy stored per unit volume in a solenoid
$U=\frac{B^2}{2{\mu }_0},B={\mu }_{0}nI$
(Magnetic field inside the solenoid)
$\therefore U=\frac{B^2}{2{\mu }_0}=\frac{{\left({\mu }_{0}nI\right)}^2}{2{\mu }_0}$
$=\frac{{\mu }_0^2{n}^2{I}^2}{2{\mu }_0}=\frac{{\mu }_0{n}^2{I}^2}{2}$
$=\frac{4\pi \times 10^{-7}\times \left(40\times 10^2\right)\times (1{)}^2}{2}$
$=3.2\pi J{m}^{-3}$