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A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\vec{F}$, the net force on the remaining three arms of the loop is
A
$\vec{F}$
B
$3\vec{F}$
C
$-\vec{F}$
D
$-3\vec{F}$
Detailed Solution
The net force on a closed current-carrying loop in a uniform magnetic field is zero: $\oint I\,d\vec{l}\times\vec{B} = I\left(\oint d\vec{l}\right)\times\vec{B} = 0$
Let $\vec{F}_1 = \vec{F}$ be the force on one arm and $\vec{F}_2$ the net force on the remaining three arms.
$\vec{F}_1 + \vec{F}_2 = 0$
$\vec{F}_2 = -\vec{F}_1 = -\vec{F}$
Let $\vec{F}_1 = \vec{F}$ be the force on one arm and $\vec{F}_2$ the net force on the remaining three arms.
$\vec{F}_1 + \vec{F}_2 = 0$
$\vec{F}_2 = -\vec{F}_1 = -\vec{F}$