The force per unit length on a wire carrying current of 8A making an angle of 30∘ with a uniform magnetic field of 0.15 T is

Magnetic force per-unit length,F=BILsin⁡θF/l=BIsin⁡θ=0.15×8×0.5=0.6Nm−1

The force per unit length on a wire carrying current of 8A making an angle of 30∘ with a uniform magnetic field of 0.15 T is

Magnetic force per-unit length,F=BILsin⁡θF/l=BIsin⁡θ=0.15×8×0.5=0.6Nm−1

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The force per unit length on a wire carrying current of 8A making an angle of $30^{\circ}$ with a uniform magnetic field of 0.15 T is

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1.2 N

1.02 N

0.6 N

2.4 N

answer is C.

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Magnetic force per-unit length,

$F = BILsin θ$

$F / l = BIsin θ = 0.15 \times 8 \times 0.5 = 0.6 {Nm}^{- 1}$

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The force per unit length on a wire carrying current of 8A making an angle of 30∘ with a uniform magnetic field of 0.15 T is

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The force per unit length on a wire carrying current of 8A making an angle of $30^{\circ}$ with a uniform magnetic field of 0.15 T is