A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil?

Explanation

To find the magnitude of torque experienced by the square coil in a magnetic field, we can use the formula for torque (\( \tau \)) on a current-carrying coil:

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

where:
- \( n \) = number of turns in the coil
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