Question

The cross-section of an infinite cylinder is shown below. A current I uniformly distributed over its cross-section is flowing along its length. Now a cylinder of radius R/2 is checked out of this. Then

Answer 1

We can consider the given cylinder as a combination of two cylinders −

One complete solid cylinder of radius R with current I in upward direction

And another solid cylinder of radius R/2 which earlier filled the hollow part and carrying current in opposite direction with current density same as solid cylinder

Current Density in larger cylinder,

$i=\frac{I}{A_{largecyl.}}$

$\Rightarrow i=\frac{I}{\pi R^2}$

Then current in hollow part cylinder,
$I_0=i\left(\frac{\pi R^2}{4}\right)$

$\Rightarrow I_0=\frac{I}{\pi R^2}\times \left(\frac{\pi R^2}{4}\right)$

$\Rightarrow I_0=\frac{I}{4}$

Magnetic field of a solid cylinder
$\overrightarrow{B}$ for a solid cylinder carrying current I at distance r < R (inside the cylinder)

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

For outside the cylinder when $r\le R$

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi r}$

r = distance of point from axis,
R = radius of cylinderical wire or conductor

$\overrightarrow{B}$ at point A
${\overrightarrow{B}}_A={\overrightarrow{B}}_1+{\overrightarrow{B}}_2$

( ${\overrightarrow{B}}_1$ is due to full cylinder and & ${\overrightarrow{B}}_2$ is due to hollow cylinder )

${\overrightarrow{B}}_A=0+\frac{{\mu }_0\left(I/4\right)}{2\pi \left(R/2\right)}=\frac{{\mu }_{0}I}{4\pi r}(\times )$

Direction can be found using maxwell's right hand thumb rule, for downward current in smaller cylinder, ${\overrightarrow{B}}_2$ is inward (taken +ve)

${\overrightarrow{B}}_1=0$, as point A lies on axis of full cylinder, r = 0 for this cylinder

Point A lies on surface of smaller cylinder so r = R/2 for this cylinder

$\overrightarrow{B}$ at point B
${\overrightarrow{B}}_B={\overrightarrow{B}}_1+{\overrightarrow{B}}_2$

(${\overrightarrow{B}}_1$ is due to full cylinder and ${\overrightarrow{B}}_2$ is due to hollow cylinder )

${\overrightarrow{B}}_B\frac{{\mu }_0\left(\frac{R}{2}\right)I}{2\pi R^2}+0=\frac{{\mu }_{0}I}{4\pi r}(\times )$

${\overrightarrow{B}}_2$ = 0, as point B lies on axis of smaller cylinder, r = 0
Point B lies at distance R/2 from axis of bigger full cylinder and it lies inside the cylinder, r = R/2

${\overrightarrow{B}}_B=\frac{{\mu }_{0}i}{4\pi r}(\times )$

Directions can be found out by maxwell's right hand thumb rule, for upward current in bigger cylinder ${\overrightarrow{B}}_1$ is inward (taken +ve)

So, magnetic fields at the centre of both cylinders is same both in magnitude and direction.

Hence, option (A) is correct.