If the solenoid is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25 T$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{o}$ with the direction of applied field? Given that magnetic moment is $0.6 J T^{- 1}$ .

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Solution

Given,

Magnetic field strength, $B = 0.25 T$

Magnetic moment, $m = 0.6 J T^{- 1}$

The angle between the axis of the solenoid and the direction of the applied field, $θ = 30 ° .$

We know, the torque acting on the solenoid is:

$τ = m x B = m B s i n θ$
$= (0.6 J T^{- 1}) (0.25 T) (s i n 30^{0})$

$= 0.075 J$
$= 7.5 \times 10^{- 2} J$

The magnitude of torque is $7.5 \times 10^{- 2} J .$

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