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 outside the solid conductor such that $R_1<r<R_2$ will be

• A. $\frac{{\mu }_0(I_1+I_2)}{2\pi r}$
• B. $\frac{{\mu }_0(I_1-I_2)}{2\pi r}$
• C. $\frac{{\mu }_0{I}_1}{2\pi R_1^2}{r}_2$
• D. $\frac{{\mu }_0{I}_1}{2\pi r}$

Answer 1

The correct option is D $\frac{{\mu }_0{I}_1}{2\pi r}$

Considering amperian loop (2)
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_{enc}B\left(2\pi r\right)={\mu }_0{I}_{1}B=\frac{{\mu }_0{I}_1}{2\pi r}$