Question

A very long cylindrical wire is carrying a current $I_o$ distributed uniformly over its cross-section area. $O$ is the centre of the cross-section of the wire and the direction of current in into the plane of the figure. The value of ${\int }_A^B\overrightarrow{B}.\overrightarrow{\mathrm{dI}}$ along the path $AB$ (from $A$ to $B$ ) is

• A. ${\mu }_0{I}_0$
• B. $-\frac{{\mu }_0{I}_0}{6}$
• C. $\frac{{\mu }_0{I}_0}{6}$
• D. $\frac{{\mu }_0{I}_0}{3}$

Answer 1

Answer: (B)

According to Ampere's circuital law
$\oint \overrightarrow{B}.\overrightarrow{\mathrm{dl}}={\mu }_0{I}_{net}$
For the closed path $OAB$
$\oint \overrightarrow{B}.\overrightarrow{\mathrm{dl}}={\int }_O^A\overrightarrow{B}.\overrightarrow{\mathrm{dl}}+{\int }_A^B\overrightarrow{B}.\overrightarrow{\mathrm{dl}}+{\int }_B^O\overrightarrow{B}.\overrightarrow{\mathrm{dl}}={\left(\mu \right)}_0\left(-\frac{I_0}{6}\right)$
Since the magnetic field is perpendicular to the paths $OA$ and $BO$
${\int }_A^B\overrightarrow{B}.\overrightarrow{\mathrm{dl}}=-\frac{{\mu }_0{I}_0}{6}$