Question

A hollow cylindrical wire carries a current $I$, having inner and outer radii $R$ and $2R$ respectively. Magnetic field at a point which $\frac{3R}{2}$ distance away from its axis is

• A. $\frac{5{\mu }_{0}I}{18\pi R}$
• B. $\frac{{\mu }_{0}I}{36\pi R}$
• C. $\frac{5{\mu }_{0}I}{36\pi R}$
• D. $\frac{5{\mu }_{0}I}{9\pi R}$

Answer 1

The correct option is C $\frac{5{\mu }_{0}I}{36\pi R}$

For a hollow cylinder having inner radius $a$ and outer radius $b$, the magnetic field in between the cylinder at distance $r(a<r<b)$ is obtained by ampere's law.

Here dotted circle is a amperian loop, so

$I_{en}=\frac{I}{A}\times A^{\prime }=\frac{I}{\pi (b^2-a^2)}\times \pi (r^2-a^2)$

$I_{en}=(\frac{r^2-a^2}{b^2-a^2})I$

Apply Ampere's circuital law,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0{I}_{en}$

$B\cdot 2\pi r={\mu }_0(\frac{r^2-a^2}{b^2-a^2})I$

$B=\frac{{\mu }_{0}I}{2\pi r}\left(\frac{r^2-a^2}{b^2-a^2}\right)$

Here,

$a=R,b=2R,r=3R/2$

$B=\frac{{\mu }_{0}I}{2\pi \times \frac{3R}{2}}\times \left(\frac{\frac{9R^2}{4}-R^2}{4R^2-R^2}\right)$

$B=\frac{{\mu }_{0}I}{3\pi R}\times \left(\frac{\frac{5R^2}{4}}{3R^2}\right)$

$\therefore B=\frac{5}{36}\frac{{\mu }_{0}I}{\pi R}$

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