A straight solenoid of length 50 cm has 1000 turns per metre and mean cross-sectional area of `2 xx 10^(-4) m^(2)` . It is placed with its axis of `30^(@)` with uniform magnetic field of 0.32 T . Find the torque acting on the solenoid when a current of 2 ampere is passed through it .

Explanation

To find the torque acting on the solenoid when a current of 2 amperes is passed through it, we can use the formula for the torque (\(\tau\)) acting on a solenoid in a magnetic field:

\[ \tau = n \cdot A \cdot B \cdot \sin(\theta) \]

Where:
- \(n\) = number of turns per unit length (in turns/m)
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