Question

A solid cylindrical conductor of radius $R_1$ is kept inside a hollow cylindrical conductor of inner radius $R_2$ and outer radius $R_3(R_1<R_2<R_3)$ in a co - axial manner, as shown. The solid cylinder carries a current $I_1$ and the hollow conductor carries current $I_2$ in the opposite directions. The two conductors are kept insulated from each other and current through each is distributed uniformly over the cross–section. Let r be any radial distance from the axis.

Magnitude of magnetic induction B at points inside the solid conductor, so that $r<R_1$ , can be expressed as

• A. $\frac{{\mu }_0{I}_1}{2\pi R_1}r$
• B. $\frac{{\mu }_0{I}_1}{2\pi R_1^2}$
• C. $\frac{{\mu }_0{I}_1}{R_1}r$
• D. $zero$

Answer 1

The correct option is B $\frac{{\mu }_0{I}_1}{2\pi R_1^2}$

From side-on view, for $r<R_1$

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_{enc}b\int dl={\mu }_0\times \frac{I_1}{\pi R_1^2}\times \pi r^{2}B\left(2\pi r\right)={\mu }_0\times \frac{I_1}{\pi R_1^2}\times r^{2}B={\mu }_0\frac{I_{1}r}{2\pi R_1^2}$