Q. The magnetic field existing in a region is given by $\overrightarrow{B} = B_{0}\left\lbrack 1 + \frac{x}{l} \right\rbrack\widehat{k}.\ $A square loop of edge $l$ and carrying current $i$ is placed with its edges parallel to $x - y$ axis. The magnitude of the net magnetic force experienced by the loop is
A) $2B_{0}il$
B) $B_{0}i_{0}l$
C) $B_{0}il$
D) $Bil$
Explanation
(c) The magnetic force near side $ad$ is
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${\overrightarrow{F}}_{1} = B_{0}\left\lbrack 1 + \frac{a_{0}}{l} \right\rbrack\widehat{k}$
The magnetic force on wire $ab\ \ and\ cd$ is equal and opposite. The magnetic field near side be is
${\overrightarrow{F}}_{2} = B_{0}\left\lbrack 1 + \frac{a_{0} + l}{l} \right\rbrack$
The net force on the loop = $F = F_{2} = F_{1}$
$= ilB_{0}\left( 1 + \frac{l + a_{0}}{l} \right) - iB_{0}l\left\lbrack 1 + \frac{a_{0}}{l} \right\rbrack$
= $B_{0}il$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.