Question

From a cylinder of radius $R$, a cylinder of radius $R/2$ is removed. Current flowing in the remaining cylinder is $I$. Magnetic field strength is

• A. zero at point A
• B. zero at point B
• C. $\frac{{\mu }_{0}I}{3\pi R}$ at point A
• D. $\frac{{\mu }_{0}I}{3\pi R}$ at point B

Answer 1

Answer: (C), (D)

$\frac{{\mu }_{0}I}{3\pi R}$ at point B

Since current carried by the part is proportional to the area of that part, the removed part of radius $R/2$ has area $A/4$ and the remaining part of the cylinder has area $3A/4$.

Current carried by the removed part of the cylinder of radius $R/2$
$=\frac{1}{3}\times$ current carried by remaining part of the cylinder
$=I/3$

Therefore, we can consider the given cylinder as a combination of two cylinders. One of radius $R$ carrying current $4I/3$ in one direction and the other of radius $R/2$ carrying current $I/3$ in opposite direction.

For cylinder,

$B=\frac{{\mu }_{o}ir}{2\pi R^2};r<R=\frac{{\mu }_{o}i}{2\pi r};r\ge R$

At point A:

$B=0+\frac{{\mu }_o\frac{I}{3}}{2\pi \frac{R}{2}}=\frac{{\mu }_{o}I}{3\pi R}$

At point B:

$B=\frac{{\mu }_o}{2\pi R^2}\left(\frac{4I}{3}\right)\left(\frac{R}{2}\right)+0=\frac{{\mu }_{o}I}{3\pi R}$