Question

Inside an infinitely long wire, having circular cross-section, there is a hole that runs throughout the length of the wire, as shown in the diagram. A current, with current density $J$, flows through the wire over the cross-section. The magnetic field at point $P$ will be -

• A. $\frac{{\mu }_{o}aJ}{2}$
• B. $\frac{{\mu }_{o}aJ}{8}$
• C. $\frac{{\mu }_{o}aJ}{4}$
• D. ${\mu }_{o}aJ$

Answer 1

The correct option is A $\frac{{\mu }_{o}aJ}{2}$

Consider a circular closed path over the cross-section of the wire, with its centre on the axis of the wire and radius equal to $x+a$.


Applying Ampere's law to this circular path, we get, magnetic field at $P$ as,

$B_1\left[2\pi \right(a+x\left)\right]={\mu }_o\left[J\pi \right(a+x{)}^2]$

$\therefore B_1=\frac{{\mu }_{o}J(a+x)}{2}$

This is the magnetic field at point $P$, due to the current flowing through the circular cross-section of radius $a+x$ assuming there is no hole.

But, there is a circular hole of radius $x$.

So, we can assume that an opposite current having the same current density $J$ is flowing through this hole.

From similar manner to calculate magnetic field as above, we get, magnetic field at point $P$ due to current through hole as,

$B_2=\frac{{\mu }_{o}Jx}{2}$

The net magnetic field at $P$ is,

$B_{net}=B_1-B_2$

$=\frac{{\mu }_{o}J(a+x)}{2}-\frac{{\mu }_{o}Jx}{2}$

$\therefore B_{net}=\frac{{\mu }_{o}aJ}{2}$

Hence, option $\left(A\right)$ is the correct answer.