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

BIL

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$\frac{3}{2} B I L$

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zero

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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 $| \to l | = \frac{L}{2}$ and $θ = 90^{\circ} . . . . . F = I \times \frac{L}{2} \times B = \frac{B I L}{2}$

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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

The correct option is D BIL2 →F=I(→l×→B) Here →l is directed from end A to end B of the wire. Here |→l|=L2 and θ=90∘.....F=I×L2×B=BIL2

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

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 $| \to l | = \frac{L}{2}$ and $θ = 90^{\circ} . . . . . F = I \times \frac{L}{2} \times B = \frac{B I L}{2}$