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A bent wire AB carrying a current I is placed in a region of uniform magnetic field $B$ . The force on the wire AB is

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$\frac{3}{2} BIL$

BIL

$\frac{BIL}{2}$

Zero

answer is D.

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Detailed Solution

$\vec{F} = I (\vec{l} \times \vec{B})$
Here $ι$ is directed from end A to end B of the wire.
Here $| \vec{l} | = \frac{L}{2} and θ = 90^{\circ} \dots . F = I \times \frac{L}{2} \times B = \frac{BIL}{2}$

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