Question

The axis of a solid cylinder of infinite length and radius $R$ lies along $y-$ axis. It carries a uniformly distributed current $i$ along $+vey-$ direction. The magnetic field at a point $(\frac{R}{2},y,\frac{R}{2})$ is:

• A. $\frac{{\mu }_{0}i}{4\pi R}(\hat{i}-\hat{k})$
• B. $\frac{{\mu }_{0}i}{2\pi R}(\hat{j}-\hat{k})$
• C. $\frac{{\mu }_{0}i}{4\pi R}\hat{j}$
• D. $\frac{{\mu }_{0}i}{4\pi R}(\hat{i}+\hat{k})$

Answer 1

The correct option is A $\frac{{\mu }_{0}i}{4\pi R}(\hat{i}-\hat{k})$

The solid cylinder is carrying current along $y-$ axis. Thus, the magnetic field at a point lying on its axis will be zero.

For point $\text{P}(\frac{R}{2},y,\frac{R}{2})$, the magnetic field will only correspond due to its position on $x\&z-$ axis.

Since, $\frac{R}{2}<R$ , so the magnetic field at inside the region is given by

$B=\frac{{\mu }_{0}id}{2\pi R^2}$

Here, $d=R/2$

$B=\frac{{\mu }_{0}i\left(R/2\right)}{2\pi R^2}=\frac{{\mu }_{0}i}{4\pi R}$

Using right-hand thumb rule, the magnetic field at $x=\frac{R}{2}$ is

$B_x=\frac{{\mu }_{0}iR}{4\pi R^2}(-\hat{k})=-\frac{{\mu }_{0}i}{4\pi R}\hat{k}$

Similarly, the magnetic field at $z=\frac{R}{2}$ is

$B_z=\frac{{\mu }_{0}i}{4\pi R}\hat{i}$

Thus, ${\overrightarrow{B}}_{net}={\overrightarrow{B}}_x+{\overrightarrow{B}}_y+{\overrightarrow{B}}_z$

$B_{net}=-\frac{{\mu }_{0}i}{4\pi R}\hat{k}+0+\frac{{\mu }_{0}i}{4\pi R}\hat{i}$

$\therefore B_{net}=\frac{{\mu }_{0}i}{4\pi R}(\hat{i}-\hat{k})$

Therefore, option $\left(a\right)$ is correct.