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Class 12 › Physics › Moving Charges and Magnetism
CBSE2023Class Class 12 · Physics3 Marks · Short✅ Verified
A circular coil of 30 turns and radius 8.0 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of 1.0 T. The field lines make an angle of $30^\circ$ with the plane of the coil. Calculate the magnitude of the external torque needed to prevent the coil from turning. What happens if the circular coil is replaced by a planar coil of irregular shape enclosing the same area, keeping other parameters unchanged?
✅ Answer & Solution
Torque on a current-carrying coil: $$\tau = NBIA\sin\theta$$ where $\theta$ is the angle between the magnetic field and the normal to the coil. The field makes $30^\circ$ with the plane of the coil, so the angle with the normal is $\theta = 90^\circ - 30^\circ = 60^\circ$. Area: $$A = \pi r^2 = \pi(0.08)^2 = \pi(6.4\times10^{-3}) = 2.01\times10^{-2}\ \text{m}^2$$ Torque: $$\tau = (30)(1.0)(6)(2.01\times10^{-2})\sin60^\circ$$ $$= (30)(6)(2.01\times10^{-2})(0.866) \approx 3.13\ \text{N·m}$$ To prevent turning, the external torque must equal this in magnitude: $\tau_{ext}\approx3.13$ N·m. Replacing the circular coil by an irregular planar coil of the same area: the torque depends only on the area enclosed, not the shape, so the torque remains unchanged.
