Q.A short bar magnet of magnetic moment $m = 0.32\ \text{J T}^{-1}$ is placed in a uniform magnetic field of $0.15\ \text{T}$. If the bar is free to rotate in the plane of the field, which orientation would correspond to its

Concept: Magnetic Poles — A magnetic dipole in a uniform field experiences a torque that tries to align it with the field. Stable equilibrium occurs when the dipole is parallel to the field (lowest potential energy), and unstable equilibrium when it is antiparallel (highest potential energy).

Reasoning:

1. Potential energy of a magnetic dipole in a uniform field: $U = - m B \cos \theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$.
2. For stable equilibrium, $U$ is minimum → $\cos \theta = +1$ → $\theta = 0^\circ$ (parallel).
3. For unstable equilibrium, $U$ is maximum → $\cos \theta = -1$ → $\theta = 180^\circ$ (antiparallel).
4. Compute $U$: $m = 0.32\ \text{J T}^{-1}$, $B = 0.15\ \text{T}$ → $mB = 0.048\ \text{J}$.

The key idea is that a magnetic dipole — like a bar magnet — in a uniform external field experiences a torque that tries to align it with the field. But the real story is about energy. The potential energy of a magnetic dipole in a uniform field is given by $U = -\vec{m} \cdot \vec{B} = -mB\cos\theta$, where $\theta$ is the angle between the magnetic moment vector $\vec{m}$ and the field $\vec{B}$.

Why does this matter for equilibrium? Because nature always seeks the lowest energy state. When the magnet is aligned with the field ($\theta = 0^\circ$), $\cos\theta = 1$, so $U = -mB$ — the most negative, hence lowest, energy. That’s stable equilibrium: if you nudge it, it will return. When it’s anti-aligned ($\theta = 180^\circ$), $\cos\theta = -1$, so $U = +mB$ — the highest energy. That’s unstable: the slightest push sends it spinning toward the stable orientation.

Let’s work through the numbers.

1. Identify the given data

   Magnetic moment, $m = 0.32\ \text{J T}^{-1}$

   Magnetic field strength, $B = 0.15\ \text{T}$

   The magnet is free to rotate in the plane of the field, so $\theta$ can vary from $0^\circ$ to $180^\circ$.

2. Write the potential energy formula

3. Stable equilibrium This occurs at the minimum of $U$. Since $\cos\theta$ is maximum at $\theta = 0^\circ$, we have:

Substitute:

The orientation: the magnet’s north pole points in the direction of the external field.

4. Unstable equilibrium This occurs at the maximum of $U$, at $\theta = 180^\circ$:

So:

The orientation: the magnet’s north pole points opposite to the external field.

Method: Potential Energy Analysis for a Magnetic Dipole in a Uniform Field

This problem uses the potential energy method for a magnetic dipole in a uniform external field. The key idea: a system is in stable equilibrium when its potential energy is minimum, and in unstable equilibrium when its potential energy is maximum.

Step-by-step solution

Step 1: Recall the potential energy formula

For a magnetic dipole (bar magnet) of magnetic moment $\vec{m}$ placed in a uniform magnetic field $\vec{B}$, the potential energy is:

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

Step 2: Identify the equilibrium conditions

• Stable equilibrium: $U$ is minimum → $\cos\theta$ is maximum → $\cos\theta = +1$ → $\theta = 0^\circ$

  (magnetic moment aligned parallel to the field)

• Unstable equilibrium: $U$ is maximum → $\cos\theta$ is minimum → $\cos\theta = -1$ → $\theta = 180^\circ$

  (magnetic moment anti-parallel to the field)

Step 3: Calculate the potential energies

Given:

$m = 0.32\ \text{J T}^{-1}$

$B = 0.15\ \text{T}$

1. Stable equilibrium ($\theta = 0^\circ$, $\cos\theta = 1$):

   $U_{\text{stable}} = -mB\cos 0^\circ = -mB = -(0.32)(0.15)$

   $\boxed{U_{\text{stable}} = -0.048\ \text{J}}$

2. Unstable equilibrium ($\theta = 180^\circ$, $\cos\theta = -1$):

   $U_{\text{unstable}} = -mB\cos 180^\circ = -mB(-1) = +mB = (0.32)(0.15)$

   $\boxed{U_{\text{unstable}} = +0.048\ \text{J}}$

Final Answer Summary

Key insight: The negative sign in $U = -mB\cos\theta$ is crucial — it makes the aligned orientation energetically favourable (lower energy), which is why a freely rotating magnet always settles parallel to the field.

Here are the common mistakes students make on this magnetic poles / torque & potential energy question, and how to avoid each.

Mistake 1: Confusing stable and unstable equilibrium orientations

The error:

Students often think the magnet aligns perpendicular to the field for stable equilibrium, or they swap the two orientations.

Why it happens:

They memorise “stable = minimum energy” but forget the actual angular dependence.

How to avoid:

• Stable equilibrium → magnetic moment $\vec{m}$ is parallel to $\vec{B}$ (angle $\theta = 0^\circ$).
• Unstable equilibrium → $\vec{m}$ is antiparallel to $\vec{B}$ ($\theta = 180^\circ$).

Think of a compass needle: it points along the field (stable). Flipping it 180° is unstable — the slightest nudge makes it swing back.

Mistake 2: Using the wrong formula for potential energy

The error:

Using $U = -mB\cos\theta$ but forgetting the negative sign, or using $U = +mB\cos\theta$.

Why it happens:

Misremembering the sign convention.

How to avoid:

The potential energy of a magnetic dipole in a uniform field is:

• At $\theta = 0^\circ$: $\cos 0 = 1$ → $U = -mB$ (minimum → stable)
• At $\theta = 180^\circ$: $\cos 180 = -1$ → $U = +mB$ (maximum → unstable)

Check: Minimum energy = stable; maximum energy = unstable.

Mistake 3: Forgetting to include units or misreading given data

The error:

Writing $U = -0.32 \times 0.15 = -0.048$ but omitting the unit (Joules), or misreading $m$ as $0.32\ \text{J T}^{-1}$ and $B$ as $0.15\ \text{T}$.

How to avoid:

Always write the full calculation with units:

Pro tip: In exams, box your final answer with the correct unit.

Mistake 4: Thinking torque is zero only in stable equilibrium

The error:

Stating that torque is zero only for $\theta = 0^\circ$, forgetting $\theta = 180^\circ$ also gives zero torque.

Why it happens:

Torque $\tau = mB\sin\theta$ — students see $\sin 0 = 0$ but forget $\sin 180 = 0$ too.

How to avoid:

Torque is zero at both $\theta = 0^\circ$ and $\theta = 180^\circ$. The difference is:

• $\theta = 0^\circ$ → stable (restoring torque if displaced)
• $\theta = 180^\circ$ → unstable (torque away from equilibrium if displaced)

Quick Summary Table

Final tip: Draw a quick diagram — arrow for $\vec{m}$ and arrow for $\vec{B}$ — before writing the answer. It prevents orientation errors every time.