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Q.

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of $30 °$ with the direction of a uniform magnetic field of 0.15 T ?

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a

$4 N m^{- 1}$

b

$6 N m^{- 1}$

c

$0.4 {Nm}^{- 1}$

d

$0.6 N m^{- 1}$

answer is D.

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Detailed Solution

Force on a current carrying element in a magnetic field is given by

$F = B I L \sin θ$

$\frac{F}{L} = B I \sin θ$

$\Rightarrow \frac{F}{L} = 0.15 \times 8 \times \frac{1}{2} = 0.6 N / m$

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