question_answer

The magnetic field existing in a region is given by\[\overrightarrow{B}={{B}_{0}}\left[ 1+\frac{x}{\ell } \right]\widehat{k}\]. A square loop of edge\[\ell \]. and carrying current I is placed with its edges  parallel to the x-y axis. The magnitude of the net magnetic force experienced by the loop is-

A) \[2{{B}_{0}}I\ell \]

B)               \[{{B}_{0}}{{I}_{0}}\ell \]

C) \[{{B}_{0}}I\ell \]                     

D) \[BI\ell \]

Correct Answer: C

Solution :

The magnetic force near side ad is \[{{\overrightarrow{\text{F}}}_{\text{1}}}\,\text{=}\,{{\text{B}}_{\text{0}}}\left[ 1+\frac{{{a}_{0}}}{\ell } \right]\widehat{k}\] The magnetic force on wire ab and cd is equal and opposite. The magnetic field near side be is \[\overrightarrow{{{F}_{2}}}\,\,=\,{{B}_{0}}\left[ 1+\frac{{{a}_{0}}+\ell }{\ell } \right]\] The net force on the loop \[=\text{ }F={{F}_{2}}-{{F}_{1}}\] \[=\,\,I\ell {{B}_{0}}\,\left( 1+\frac{\ell +{{a}_{0}}}{\ell } \right)\,\,-I{{B}_{0}}\ell \,\,\left[ 1+\frac{{{a}_{0}}}{\ell } \right]\] \[=\,\,\,{{B}_{0}}I\ell \]