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

B I L

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\frac{1}{4} B I L

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Z e r o

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\frac{B I L}{2}

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Solution

The correct option is D $\frac{B I L}{2}$
$\to F = I (\to l \times \to B) .$
Here $\to l$ is directed from end A to end B of the wire.
Here $| \to l | = \frac{L}{2} a n d θ = 90^{\circ} . ∴ F = l \times \frac{L}{2} \times B = \frac{B I L}{2}$

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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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