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 \[=\text{ }F={{F}_{2}}-{{F}_{1}}\] \[=\,\,I\ell {{B}_{0}}\,\left( 1+\frac{\ell +{{a}_{0}}}{\ell } \right)\,\,-I{{B}_{0}}\ell \,\,\left[ 1+\frac{{{a}_{0}}}{\ell } \right]\] \[=\,\,\,{{B}_{0}}I\ell \]You need to login to perform this action.
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