Physics · Magnetism And Matter · NEET
Yes. Torque τ = MB sinθ. When the magnet points the same way as B, θ = 0°, and sin0° = 0, so τ = 0. The magnet feels no turning effect here — this is its rest (stable equilibrium) position. It is also zero at θ = 180° (magnet flipped, opposite to B), but that position is unstable and any small push turns it back to 0°.
Torque follows sinθ, not θ itself. sinθ is largest (= 1) at θ = 90°, so τ = MB is maximum when the magnet lies exactly perpendicular (sideways) to the field. At 180° the magnet is anti-parallel and sin180° = 0, so torque is again zero. Students often wrongly think 'more angle means more torque' — track the sine value, not the raw angle.
In a UNIFORM field the net force on a magnet is zero (the pull on the N pole and the equal push on the S pole cancel), so the magnet does not move to one side. But these two equal-opposite forces act at different points, forming a couple, so they still produce a torque τ = MB sinθ that rotates the magnet. Net force zero, net torque not zero. A NON-uniform field is needed to get a net translational force.
Torque uses sinθ: τ = MB sinθ. The cosθ form belongs to potential energy, U = −MB cosθ. Easy check: torque must vanish when the magnet is aligned (θ = 0), and sin0° = 0 gives that. If you wrongly used cos, you'd get τ = MB at θ = 0, which is impossible for an aligned magnet.
Only rotates it. A uniform field has the same strength and direction everywhere, so the force on the N pole (+qm B) and on the S pole (−qm B) are equal and opposite and cancel — no net pull. The magnet just aligns itself along the field, like a compass needle turning to point North.
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium. The energy required to rotate it by 60° is W. The torque required to keep the magnet in this new (60°) position is:
A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 A and is in a magnetic field of strength 0.85 T. The work done in rotating the coil by 180° against the torque is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
τ = MB sinθ, or in vector form τ = M × B, where M is the magnetic dipole moment (A·m²), B is the field (tesla), and θ is the angle between M and B. SI unit of torque is N·m.
Maximum τ = MB when the magnet is perpendicular to the field (θ = 90°). Zero when the magnet is parallel (θ = 0°, stable) or anti-parallel (θ = 180°, unstable) to the field.
Equal and opposite pole forces (+qm B and −qm B) cancel, giving zero net force, but they act at different points forming a couple, which produces torque τ = MB sinθ. Net force needs a non-uniform field.
For a flat coil, M = N I A, where N is number of turns, I is current, and A is the area of one turn. Then use τ = M B sinθ just like a bar magnet.
Torque is the tendency to align (τ = MB sinθ); potential energy is U = −MB cosθ. Energy is lowest (−MB) when aligned and highest (+MB) when anti-parallel. Work to rotate = change in U = MB(cosθ₁ − cosθ₂).
