Q.A bar magnet of magnetic moment $1.5\ \text{J T}^{-1}$ lies aligned with the direction of a uniform magnetic field of $0.22\ \text{T}$.

Concept: Magnetic Poles — the work done by an external torque to rotate a magnetic dipole in a uniform field equals the change in its potential energy, $U = - \vec{M} \cdot \vec{B}$.

Given: $M = 1.5\ \text{J T}^{-1}$, $B = 0.22\ \text{T}$.

(a) Work required = $U_f - U_i$, with initial angle $\theta_i = 0^\circ$ (aligned).

• (i) Final $\theta_f = 90^\circ$ (normal).

  $W = (-MB\cos 90^\circ) - (-MB\cos 0^\circ) = 0 + MB = (1.5)(0.22) = 0.33\ \text{J}$.

• (ii) Final $\theta_f = 180^\circ$ (opposite).

  $W = (-MB\cos 180^\circ) - (-MB\cos 0^\circ) = MB - (-MB) = 2MB = 0.66\ \text{J}$. …

Concept and Intuition

A bar magnet in a uniform magnetic field behaves like a compass needle — it experiences a torque that tries to align it with the field. But here, we are not letting it align naturally; we are using an external agent to rotate it to specific orientations. The key idea: the work done by the external torque equals the change in the magnet’s potential energy in the field.

Why? Because the magnetic field does work on the magnet as it rotates, and the external torque must oppose that to achieve the desired orientation. The potential energy of a magnetic dipole in a uniform field is given by:

where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$. The lowest energy is at $\theta = 0^\circ$ (aligned), and the highest at $\theta = 180^\circ$ (anti-aligned). So rotating away from alignment increases potential energy — that increase is the work you must supply.

Step-by-Step Solution

Given:

$M = 1.5\ \text{J T}^{-1}$, $B = 0.22\ \text{T}$

Initial orientation: aligned with the field → $\theta_1 = 0^\circ$.

1. Work required to turn the magnet

Work done by external torque = change in potential energy:

(a)(i) Normal to the field: $\theta_2 = 90^\circ$

(a)(ii) Opposite to the field: $\theta_2 = 180^\circ$

$W = 2 \times 0.33 = 0.66\ \text{J}$ …

Method: Work–Energy Relation for a Magnetic Dipole in a Uniform Field

This method uses the potential energy of a magnetic dipole in a uniform field and the torque expression directly.

Key formulas

• Potential energy of a magnetic dipole:

  $U = - \vec{M} \cdot \vec{B} = - M B \cos \theta$

  where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

• Work done by an external agent to rotate the magnet:

  $W = \Delta U = U_f - U_i$

• Torque on the dipole:

  $\tau = M B \sin \theta$

Step-by-step solution

Given:

$M = 1.5\ \text{J T}^{-1}$, $B = 0.22\ \text{T}$

Initial angle: $\theta_i = 0^\circ$ (aligned with field)

(a) Work required

(i) Final angle $\theta_f = 90^\circ$ (normal to field)

• Initial energy: $U_i = - M B \cos 0^\circ = - M B$
• Final energy: $U_f = - M B \cos 90^\circ = 0$
• Work done: $W = U_f - U_i = 0 - (- M B) = M B$

$W = 1.5 \times 0.22 = 0.33\ \text{J}$

(ii) Final angle $\theta_f = 180^\circ$ (opposite to field)

• Initial energy: $U_i = - M B$
• Final energy: $U_f = - M B \cos 180^\circ = - M B (-1) = + M B$
• Work done: $W = U_f - U_i = M B - (- M B) = 2 M B$ …

Here are the common mistakes students make on this magnetic pole / dipole problem, and how to avoid each.

Mistake 1: Confusing Work Done by External Torque with Work Done by the Field

The error:

Students often use the formula $W = -\vec{\tau} \cdot \Delta \theta$ or think the work done by the external agent is the same as the work done by the magnetic field.

Why it’s wrong:

The magnetic field does negative work when the dipole rotates from aligned to anti-aligned. The external torque does the opposite (positive) work to overcome the field’s torque.

How to avoid:

Always use the potential energy approach for work done by an external agent:

Work done by external torque = change in potential energy:

For (a)(i):

Initial: $\theta_i = 0^\circ$, $U_i = -MB$

Final: $\theta_f = 90^\circ$, $U_f = 0$

So $W = 0 - (-MB) = MB$

For (a)(ii):

Final: $\theta_f = 180^\circ$, $U_f = +MB$

So $W = MB - (-MB) = 2MB$

Key takeaway: $W_{\text{ext}} = \Delta U$, not $-\Delta U$.

Mistake 2: Using the Wrong Sign for Torque

The error:

Writing $\tau = MB\sin\theta$ but forgetting that torque is a vector — or using $\tau = MB\cos\theta$ by mistake.

Why it’s wrong:

Torque magnitude is $|\vec{\tau}| = MB\sin\theta$, where $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

How to avoid:

Memorise the cross product: $\vec{\tau} = \vec{M} \times \vec{B}$.

Magnitude: $\tau = MB\sin\theta$.

For (b)(i): $\theta = 90^\circ$, $\sin 90^\circ = 1$ → $\tau = MB$

For (b)(ii): $\theta = 180^\circ$, $\sin 180^\circ = 0$ → $\tau = 0$

Check: At $\theta = 0^\circ$ or $180^\circ$, the dipole is aligned or anti-aligned — torque is zero.

Mistake 3: Forgetting Units or Mixing Them Up

The error:

Writing $M = 1.5\ \text{J T}^{-1}$ but then treating it as if it were in $\text{A m}^2$ without converting — or reporting work in $\text{J T}^{-1}$.

Why it’s wrong:

$1\ \text{J T}^{-1} = 1\ \text{A m}^2$, so the numerical value is fine. But the unit of work must be joules (J), not $\text{J T}^{-1}$.

How to avoid:

Always check final units:

$W = MB$ → units: $(\text{J T}^{-1}) \times (\text{T}) = \text{J}$ ✓

$\tau = MB$ → units: $(\text{J T}^{-1}) \times (\text{T}) = \text{J}$ (but torque is in N m, which is dimensionally same as J) ✓

Mistake 4: Not Distinguishing Between “Normal to” and “Opposite to”

The error:

Thinking “normal to” means $\theta = 0^\circ$ or $180^\circ$, or mixing up the two cases.

How to avoid: …