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
$\vec{B}_{N}=\frac{1}{2} \mu_{0} \left(\vec{J} \times\overrightarrow{OM}\right)-\frac{1}{2} \mu_{0} \left(\vec{J}\times\overrightarrow{PM}\right)$
$= \frac{1}{2}\mu_{0} \vec{j} \times\left(\overline{OM} -\overline{PM}\right) = \frac{1}{2}\mu_{0} \vec{j} \times\vec{d}$