Question

A current of $1.0\text{A}$ is established in a tightly wound solenoid of radius $2\text{cm}$ having $1000\text{turns/meter}.$ Find the magnetic energy stored in each meter of the solenoid.

• A. $1.6\times {10}^{-3}\text{J}$
• B. $3.2\times {10}^{-4}\text{J}$
• C. $4.5\times {10}^{-3}\text{J}$
• D. $7.9\times {10}^{-4}\text{J}$

Answer 1

The correct option is D $7.9\times {10}^{-4}\text{J}$

Given:
$i=1\text{A};r=2\text{cm}=0.02\text{m}$
$n=1000\text{turns/m}$

Magnetic field inside the solenoid is,

$B={\mu }_{0}ni$

$\therefore$ Magnetic energy stored in each meter the solenoid is,

$E=u\times \text{volume of solenoid of length 1 m}$

Here, $u=$ energy density

$E=\left(\frac{B^2}{2{\mu }_0}\right)\times \pi r^{2}l=\frac{({\mu }_{0}ni{)}^2}{2{\mu }_0}\times \pi r^{2}l$

Substituting the values,

$E=\frac{(4\pi \times {10}^{-7}\times {10}^3\times 1{)}^2}{2\times 4\pi \times {10}^{-7}}\times \pi \times (2\times {10}^{-2}{)}^2\times 1[\because l=1\text{m}]$

$E=7.9\times {10}^{-4}\text{J}$

Hence, $\left(\text{D}\right)$ is the correct answer.