Q. A small circular flexible loop of wire of radius $r$ carries a current $I$. It is placed in a uniform magnetic field $B$. The tension in the loop will be doubled if
A) $I$ is doubled
B) $B$ is halved
C) $r$ is doubled
D) Both $B$ and $I$ are doubled
Explanation
(a) $F = MB,\sin\theta$
$F = \left( I.\pi r^{2} \right).B\sin\theta$
$\frac{F_{1}}{F_{2}} = \frac{I_{1}B_{1}}{I_{2}.B_{2}} \Rightarrow \frac{F}{2F} = \frac{I.B}{I_{2}.B} \Rightarrow I_{2} = 2I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.