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A loop carrying current in a magnetic field behaves like a small compass needle. The field twists it until its magnetic moment points along the field, so turning it any other way takes work. Most board and JEE questions on this come down to one line: W = mB(cos θ1 − cos θ2).
What formula gives the work done to rotate a current loop?
Start with three results from OpenStax University Physics, Volume 2, section 11.5.
The magnetic dipole moment of a coil of N turns, each of area A, carrying current I, is m = NIA, pointing along the normal to the coil's plane.
The torque on the loop in a uniform field B is τ = m × B, so its size is mB sin θ.
The potential energy of the loop in the field is U = −m · B = −mB cos θ.
Here θ is the angle between the magnetic moment (the normal to the coil) and the field. If you turn the loop slowly, so it gains no kinetic energy, the work you do goes entirely into potential energy:
W = U2 − U1 = (−mB cos θ2) − (−mB cos θ1) = mB(cos θ1 − cos θ2).
Integrating the torque gives the same answer: the integral of mB sin θ dθ from θ1 to θ2 is mB(cos θ1 − cos θ2). For the torque formula built from the forces on each side, see our derivation of the torque on a rectangular coil.
Why does turning from 0° to 180° need the most work?
The work depends only on cos θ1 − cos θ2. Cosine runs from +1 at 0° to −1 at 180°, so the biggest possible difference is 1 − (−1) = 2. The maximum work is therefore 2mB.
OpenStax notes that a loop turns to align with the field to minimise its potential energy. From U = −mB cos θ, 0° is the bottom of the energy curve (−mB) and 180° is the top (+mB). At both angles the torque mB sin θ is zero, but they are very different: at 0° a small nudge brings the loop back (stable equilibrium), while at 180° a small nudge lets it swing all the way round (unstable equilibrium). The electric dipole works the same way, as we explain in why 0° is stable and 180° is not for a dipole in a field.
A worked example: one coil, five turns of the wrist
Take a square coil of 50 turns, each 10 cm by 10 cm (A = 0.01 m²), carrying 2 A in a uniform field of 0.2 T. Using m = NIA from OpenStax section 11.5:
m = 50 × 2 × 0.01 = 1 A m², so mB = 1 × 0.2 = 0.2 J.
Now apply W = mB(cos θ1 − cos θ2) to five different turns:
From 0° to 90°: W = 0.2(1 − 0) = 0.2 J.
From 90° to 180°: W = 0.2(0 − (−1)) = 0.2 J.
From 0° to 180°: W = 0.2(1 − (−1)) = 0.4 J, the maximum, 2mB.
From 60° to 180°: W = 0.2(0.5 − (−1)) = 0.3 J.
From 180° back to 0°: W = 0.2(−1 − 1) = −0.4 J. The answer is negative because the field does the work; you have to hold the coil back to stop it speeding up.
Two checks: steps 1 and 2 add up to step 3, because the work depends only on the start and end angles; and no answer can exceed 2mB.
What mistakes cost marks in these questions?
Using the angle with the coil's plane instead of with its normal.
Mixing up the work you do with the work the field does. They are equal and opposite.
Forgetting N. A 50-turn coil has 50 times the moment of a single turn.
Multiplying the torque at one angle by the angle turned. The torque is not constant.
Units: m is in A m², B in tesla, W in joules.
How should you practise current-loop energy problems?
Mix the turns: one from 0°, one ending at 90°, one back towards the field, and one with the angle given to the plane. When you get one wrong and cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the line where your angle or sign slipped. If the idea of magnetic moment itself feels shaky, the Fermi Concept Library gives each physics concept its own lesson and its place in the map.
Try the five turns above again tomorrow without looking, and check each answer against 2mB.
