Question

A long straight wire of circular cross-section is made of a non-magnetic material. The wire is of radius $25\text{cm}$. The wire carries a current of $10\text{A}$ which is uniformly distributed over its cross-section. The energy stored per unit length in the magnetic field contained within the wire is :

• A. $5\times {10}^{-6}\text{A}$
• B. $2.5\times {10}^{-6}\text{A}$
• C. ${10}^{-6}\text{A}$
• D. $7\times {10}^{-6}\text{A}$

Answer 1

The correct option is B $2.5\times {10}^{-6}\text{A}$

Consider a cylinder of radius $r$, thickness $dr$ and length $1\text{m}$ as shown in the figure.


From the Ampere's circuital law, the magnetic field at distance $r$ from the axis of cylinder is,

$\oint B.dl={\mu }_{0}i$

Here $i$ is the total current passing through the cylinder of radius $r$.

$\Rightarrow B\times 2\pi r={\mu }_0\frac{I}{\pi a^2}\times \pi r^2$

$\Rightarrow B=\frac{{\mu }_{0}Ir}{2\pi a^2}$

$\therefore$ Magnetic energy linked with the cylinder of thickness $dr$ is,

$dU=\frac{B^2}{2{\mu }_0}\times \text{Volume}$

Volume of cylinder of thickness $dr$,

$\Rightarrow dU={\left(\frac{{\mu }_{0}Ir}{2\pi a^2}\right)}^2\times \frac{1}{2{\mu }_0}\times 2\pi r\times h\times dr$

$=\frac{{\mu }_0{I}^2}{4\pi a^4}{r}^{3}dr[\because h=1\text{m}]$

The total magnetic energy linked with the cylindrical wire per unit its length is,

$U={\int }_0^{a}dU={\int }_0^a\frac{{\mu }_0{I}^2}{4\pi a^4}{r}^{3}dr$

$=\frac{{\mu }_0{I}^2}{4\pi a^4}{\left[\frac{r^4}{4}\right]}_0^a=\frac{{\mu }_0{I}^2}{16\pi }$

Using the data given in the question,

$U=\frac{4\pi \times {10}^{-7}\times {10}^2}{16\pi }=2.5\times {10}^{-6}\text{J/m}$

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