Question

A hollow cylindrical tube of a metal with inner and outer radii as $r_1$ and $r_2$ respectively carries a current $I$. The magnetic field inside the tube at a distance $r$ (shown in the diagram) from the centre will be

• A. $\frac{{\mu }_{o}I(r-r_1)}{2\pi r(r_2-r_1)}$
• B. $\frac{{\mu }_{o}I(r+r_1)}{2\pi r(r_2+r_1)}$
• C. $\frac{{\mu }_{o}I(r^2-r_1^2)}{2\pi r(r_2^2-r_1^2)}$
• D. $\frac{{\mu }_{o}I(r^2+r_1^2)}{2\pi r(r_2^2+r_1^2)}$

Answer 1

The correct option is C $\frac{{\mu }_{o}I(r^2-r_1^2)}{2\pi r(r_2^2-r_1^2)}$

Consider a closed path in the form of a circle of radius $r$.

If $I^{\prime }$ is the current flowing through the circle of radius $r$, then using Ampere's law, we have

$B\left(2\pi r\right)={\mu }_o{I}^{\prime }$

$\therefore B=\frac{{\mu }_o{I}^{\prime }}{2\pi r}...\left(i\right)$

Now, the current per unit area will be the same for all cross-sections.

$\therefore J=\frac{I}{\pi (r_2^2-r_1^2)}=\frac{I^{\prime }}{\pi (r^2-r_1^2)}$

$\Rightarrow I^{\prime }=\frac{I(r^2-r_1^2)}{(r_2^2-r_1^2)}...\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$B=\frac{{\mu }_{o}I(r^2-r_1^2)}{2\pi r(r_2^2-r_1^2)}$

Hence, option $\left(c\right)$ is the correct answer.