Question

Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius ${}^{\prime }{R}^{\prime }$ and outer radius ${}^{\prime }2R^{\prime }$, cylinder carrying uniformly distributed current $i$ along its axis. The magnetic induction at point ${}^{\prime }{P}^{\prime }$ at a distance $3R/2$ from the axis of the cylinder will be

• A. Zero
• B. $\frac{5{\mu }_{0}i}{72\pi R}$
• C. $\frac{7{\mu }_{0}i}{18\pi R}$
• D. $\frac{5{\mu }_{0}i}{36\pi R}$

Answer 1

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

By using the formula for magnetic field due to cylindreical conductor:

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

here $r=\frac{3R}{2},a=R,b=2R$

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

$=\frac{5.{\mu }_{0}i}{36\pi R}$