Question

Inside a long straight uniform wire of round cross-section, there is a long round cylindrical cavity whose axis is parallel to the axis of the wire and displaced from latter by a distance d. If a direct current of density $\overline{J}$ flows along the wire, then magnetic field inside the cavity will be:

• A. $0$
• B. $\frac{1}{2}{\mu }_0\overline{J}\times \overline{d}$
• C. ${\mu }_0\overline{J}\times \overline{d}$
• D. $\frac{3}{2}{\mu }_0\overline{J}\times \overline{d}$

Answer 1

The correct option is B $\frac{1}{2}{\mu }_0\overline{J}\times \overline{d}$

From Amperes law, $\oint \overline{B}.\overline{dl}={\mu }_0\int \overline{J}.\overline{ds}$

Inside the conductor at a distance r from its axis
$B\left(2\pi r\right)={\mu }_{0}J\left(\pi r^2\right);$ B=$\frac{1}{2}{\mu }_{0}Jr$

Vectorially,B=$\frac{1}{2}{\mu }_0\overline{J}\times \overline{r}$

Now, conductor has a cavity so we can assume that current in it is the superposition of positive current and negative current (in place of cavity).
Required magnetic field at point M
${\overline{B}}_n$ = Magnetic field at M due to whole conductor -
Magnetic field at M due to cavity shaped conductor

${\overline{B}}_n$=$\frac{1}{2}{\mu }_0(\overline{J}\times \overline{OM})-\frac{1}{2}{\mu }_0(\overline{J}\times \overline{PM})$

$=\frac{1}{2}{\mu }_0(\overline{J}\times \overline{OM}-\overline{PM})=\frac{1}{2}{\mu }_0\overline{J}$ X $\overline{d}.$