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A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is $\vec{F}$, the force on the arm AC is
A
$\sqrt{2}\vec{F}$
B
$-\sqrt{2}\vec{F}$
C
$-\vec{F}$
D
$\vec{F}$
Detailed Solution
Force on a current-carrying straight conductor: $\vec{F} = I\vec{L}\times\vec{B}$
The net force on a closed current loop in a uniform magnetic field is zero: $\vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{CA} = 0$
The field is along AB, so for the arm AB the length vector is parallel to $\vec{B}$ and $\vec{F}_{AB} = 0$.
Therefore $\vec{F}_{BC} + \vec{F}_{CA} = 0$
$\vec{F}_{AC} = -\vec{F}_{BC} = -\vec{F}$
The net force on a closed current loop in a uniform magnetic field is zero: $\vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{CA} = 0$
The field is along AB, so for the arm AB the length vector is parallel to $\vec{B}$ and $\vec{F}_{AB} = 0$.
Therefore $\vec{F}_{BC} + \vec{F}_{CA} = 0$
$\vec{F}_{AC} = -\vec{F}_{BC} = -\vec{F}$