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 a solenoid placed in a magnetic field, we can use the formula for torque (\( \tau \)) on a current-carrying loop in a magnetic field:

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

Where:
- \( n \) = number of turns in the solenoid
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