Potential Energy of a Magnet in a Magnetic Field

A magnet in an external magnetic field experiences a torque that tends to align it with the field. If an external agent rotates the magnet against this torque, it must do work.

That work is stored as the magnet's magnetic potential energy.

A compass needle is a small magnetic dipole. When disturbed, it turns back toward the north–south direction. This happens because its aligned position has the lowest magnetic potential energy.

Key idea: A magnetic dipole naturally tends to move toward the orientation of minimum potential energy.

The magnetic potential energy of a magnetic dipole is the energy it possesses due to its orientation in an external magnetic field.

For a magnetic dipole of moment \[\vec m\] placed in a uniform magnetic field \[\vec B\]: U = −\[\vec m\] . \[\vec B\]

or U = −mB cos⁡ θ

where θ is the angle between \[\vec m\] and \[\vec B\].

The torque on a dipole in a uniform magnetic field is: τ = mB sin ⁡θ

Step 1: Torque acting on the magnetic dipole

Step 2: Small work done against torque

Suppose an external agent rotates the dipole slowly through a small angle dθ against the magnetic torque.

Step 3: Work done from θ1 to θ2

Step 4: Relation between work and potential energy

For slow rotation, W_ext = ΔU = U₂ − U₁

Choose U = 0 at θ = 90°. Then:

Therefore, U = −mB cos⁡ θ

Since \[\vec m\] . \[\vec B\] = mBcos⁡θ, U = −\[\vec m\] . \[\vec B\]

Question

A bar magnet of magnetic moment 1.5 J T⁻¹ is initially aligned with a uniform magnetic field of magnitude 0.22 T. Find:

1. the work required to rotate it to 90°;
2. the work required to rotate it to 180°; and
3. the torque in both final positions.

Given: 

(a) Rotation from 0° to 90°

(b) Rotation from 0° to 180°

(c) Torque at 90°

(d) Torque at 180°