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 $\vec{J}$ flows along the wire, then magnetic field inside the cavity will be

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

Answer 1

Answer: (B)

From Ampere’s law, $\oint \vec{B}\cdot d\vec{l}={\mu }_0\int \vec{J}\cdot \overrightarrow{ds}$
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, $\vec{B}=\frac{1}{2}{\mu }_0\vec{J}\times \vec{r}$
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 $M$ ${\vec{B}}_N$ =Magnetic field at $M$ due to whole conductor - Magnetic field at $M$ due to cavity shaped conductor
${\vec{B}}_N=\frac{1}{2}{\mu }_0\left(\vec{J}\times \overrightarrow{OM}\right)-\frac{1}{2}{\mu }_0\left(\vec{J}\times \overrightarrow{PM}\right)$
$=\frac{1}{2}{\mu }_0\vec{j}\times \left(\overline{OM}-\overline{PM}\right)=\frac{1}{2}{\mu }_0\vec{j}\times \vec{d}$