Question

A long solenoid with $40$ turns per cm carries a current of $1\,A$. 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

The Correct Option is A

$n=40$ turns $cm ^{-1}$
$=40 \times 10^{2}$ turns $m ^{-1} $
$I =1.00\, A$
The magnetic energy stored per unit volume in a solenoid
$U=\frac{B^{2}}{2 \mu_{0}}, \,\,B=\mu_{0} n I$
(Magnetic field inside the solenoid)
$\therefore U =\frac{B^{2}}{2 \mu_{0}}=\frac{\left(\mu_{0} n I\right)^{2}}{2 \mu_{0}}$
$=\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\, Jm ^{-3} $

Answer 2

\(n=\frac{40}{10^{-2}}\) = 4 x 10³ 

long solenoid = μ₀nI Energy per unit volume

Therefore Correct option is ‘A’ is 3.2π