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  • A long solid cylindrical straight conductor carries a current i0. Find the magnetic energy stored per meter inside it. <img alt="" src="https://cdn.entrance360.c

A long solid cylindrical straight conductor carries a current i0. Find the magnetic energy stored per meter inside it.

Option: 1

\frac{\mu_0 \mathrm{i}_0^2}{2 \pi}


Option: 2

\frac{\mu_0 \mathrm{i}_0^2}{4 \pi}


Option: 3

\frac{\mu_0 \mathrm{i}_0^2}{8 \pi}


Option: 4

\frac{\mu_0 \mathrm{i}_0^2}{16 \pi}


Answers (1)

best_answer

The current enclosed within the shaded circle

\mathrm{I}=\frac{\mathrm{i}_0}{\pi \mathrm{R}^2} \pi \mathrm{r}^2=\frac{\mathrm{i}_0 \mathrm{r}^2}{\mathrm{R}^2}

Applying Ampere’s law for the dotted loop we obtain

\begin{aligned} & \oint \mathrm{B} \cdot \mathrm{d} \ell=\mu_0 \mathrm{i} \\ \Rightarrow & \mathrm{B}(2 \pi \mathrm{r})=\mu_0\left(\frac{\mathrm{i}_0 \mathrm{r}^2}{\mathrm{R}^2}\right) \\ \Rightarrow & \mathrm{B}=\frac{\mu_0 \mathrm{i}_0 \mathrm{r}}{2 \pi \mathrm{R}^2} \end{aligned}

Energy stored in the elementary volume \mathrm{dv}(=2 \pi \mathrm{r} \ell \mathrm{dr}) is given as,

\mathrm{d U=\frac{B^2}{2 \mu_0} d v}

 The total energy stored in the conductor

\begin{aligned} & \mathrm{ =U=\int \frac{B^2}{2 \mu_0} d v }\\ & \mathrm{ \Rightarrow U=\frac{1}{2 \mu_0} \int_0^R\left(\frac{\mu_0 i_0 r}{2 \pi R^2}\right)^2 \quad(2 \pi r \ell d r) }\\ & \mathrm{\Rightarrow U=\frac{\mu_0 i_0^2 \ell}{4 \pi R^4} \int_0^R r^3 d r }\\ & \mathrm{ \Rightarrow U=\frac{\mu_0 \mathrm{i}_0^2 \ell}{4 \pi} \times \frac{1}{4}=\frac{\mu_0 \mathrm{i}_0^2}{16 \pi} \quad(\because \ell=1)} \end{aligned}

Posted by

Kuldeep Maurya

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