Physics · Magnetism And Matter · NEET
No. Torque is a turning force, tau = mB sin theta (unit N.m). Work is energy spent turning it, W = mB(cos theta1 - cos theta2) (unit joule). Torque changes with angle, so you must integrate torque over the angle to get work: W = integral of tau dtheta. A question can give you W and ask for torque at some angle, or the reverse - never mix the two formulas.
Because W = U(final) - U(initial) and U = -mB cos theta. So W = (-mB cos theta2) - (-mB cos theta1) = mB(cos theta1 - cos theta2). Put the START angle in the first cos and the END angle in the second cos. If you swap them you get the wrong sign, which is the most common NEET mistake here.
From 0 to 90: W = mB(cos 0 - cos 90) = mB(1 - 0) = mB. From 0 to 180: W = mB(cos 0 - cos 180) = mB(1 - (-1)) = 2mB. Turning it fully around (to anti-aligned) needs twice the work of turning it to 90 degrees - not the same, so read the final angle carefully.
No. The magnetic potential energy U = -mB cos theta depends only on the angle, so W depends only on the start and end angles, not on how you get there. The magnetic force here is conservative for this slow rotation, exactly like lifting a mass depends only on height change.
It means you rotate so gently that kinetic energy stays zero at every instant. Then all your applied work goes into potential energy, so W(applied) = Delta U. If the magnet were released and swung freely, some energy would become kinetic and this simple formula would not give the applied work.
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 degrees is W. The torque required to keep the magnet in this new (60 degree) 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 degrees against the torque is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
W = mB(cos theta1 - cos theta2), where m is magnetic moment, B is field strength, theta1 is the starting angle and theta2 is the final angle, both measured from the field direction.
From theta = 0 (stable) to theta = 180 (unstable): W = mB(cos 0 - cos 180) = 2mB. This is the maximum possible work for a full flip.
Rotating away from alignment (increasing angle from 0) needs positive work because you oppose the field. Rotating toward alignment releases energy, giving negative applied work.
Joule (J). Here m is in A.m^2, B in tesla (T), and their product times a dimensionless cosine gives energy in joules.
W = U2 - U1 = Delta U, where U = -mB cos theta. The work you do against the field is stored as change in magnetic potential energy.
