Q.A short bar magnet placed with its axis at $30^\circ$ with a uniform external magnetic field of $0.25\ \text{T}$ experiences a torque of magnitude equal to $4.5 \times 10^{-2}\ \text{J}$. What is the magnitude of magnetic moment of the magnet?

Concept: Magnetic Poles — torque on a magnetic dipole in a uniform field depends on the magnetic moment, field strength, and the sine of the angle between them.

Step 1: The torque on a magnetic dipole is

where $M$ is the magnetic moment, $B = 0.25\ \text{T}$, and $\theta = 30^\circ$.

Step 2: Substitute the given values:

Since $\sin 30^\circ = 0.5$, this becomes

Step 3: Solve for $M$:

The key idea here is that a bar magnet behaves like a magnetic dipole — it has a north and south pole separated by a small distance, giving it a magnetic moment $\vec{M}$. When placed in an external magnetic field $\vec{B}$, the field exerts a torque that tries to align the moment with the field. The magnitude of this torque depends on three things: the strength of the moment, the strength of the field, and the angle between them.

The formula $\tau = M B \sin\theta$ is the magnetic analogue of $\tau = pE\sin\theta$ for an electric dipole in an electric field. The $\sin\theta$ factor tells you that the torque is maximum when the dipole is perpendicular to the field ($\theta = 90^\circ$) and zero when it's aligned ($\theta = 0^\circ$ or $180^\circ$). Here, the axis is at $30^\circ$ to the field, so the angle between $\vec{M}$ (which points along the axis from south to north) and $\vec{B}$ is exactly $30^\circ$.

Let's work through the numbers.

1. Write down the torque equation. For a magnetic dipole in a uniform field,

where $\tau$ is the torque magnitude, $M$ is the magnetic moment magnitude, $B$ is the field magnitude, and $\theta$ is the angle between $\vec{M}$ and $\vec{B}$.

2. Identify the given quantities.

   • $\tau = 4.5 \times 10^{-2}\ \text{J}$ (torque has units of N·m, which is the same as J)
   • $B = 0.25\ \text{T}$
   • $\theta = 30^\circ$

3. Solve for $M$.

   Rearranging the formula:

4. Plug in the values. $\sin 30^\circ = \frac{1}{2} = 0.5$, so

5. Do the division.

Method: Torque on a Magnetic Dipole in a Uniform Field

This problem uses the torque formula for a magnetic dipole (bar magnet) placed in a uniform external magnetic field.

Steps

Step 1: Recall the torque formula

The torque $\tau$ experienced by a magnetic dipole of magnetic moment $M$ placed in a uniform magnetic field $B$ at an angle $\theta$ between the dipole axis and the field is:

Step 2: Identify the given values

• $\theta = 30^\circ$
• $B = 0.25\ \text{T}$
• $\tau = 4.5 \times 10^{-2}\ \text{J}$ (Note: torque has units of N·m, which is same as J)

Step 3: Rearrange the formula for $M$

Step 4: Substitute and calculate

Step 5: Write the final answer

Key Concept Check

• Torque is maximum when $\theta = 90^\circ$ (perpendicular)
• Torque is zero when $\theta = 0^\circ$ or $180^\circ$ (parallel or antiparallel)
• The unit $\text{A·m}^2$ is equivalent to $\text{J/T}$ for magnetic moment

Here are the common mistakes students make on this exact problem, along with how to avoid each one.

1. Using the Wrong Formula for Torque

Mistake:

Students often confuse torque on a current loop ($\tau = NIAB\sin\theta$) with torque on a magnetic dipole ($\tau = MB\sin\theta$). They may also mistakenly use $\cos\theta$ instead of $\sin\theta$.

How to avoid:

• For a bar magnet (a magnetic dipole), the torque is always:

where $\theta$ is the angle between the magnetic moment vector $\vec{M}$ and the external field $\vec{B}$.

• Memorise: Torque is maximum when $\theta = 90^\circ$ (perpendicular), and zero when aligned ($\theta = 0^\circ$). This helps you remember it’s $\sin\theta$, not $\cos\theta$.

2. Misidentifying the Angle $\theta$

Mistake:

The problem says the axis is at $30^\circ$ to the field. Many students take $\theta = 30^\circ$ directly, but sometimes the angle given is between the axis and the field — which is exactly $\theta$ for a bar magnet.

How to avoid:

• For a bar magnet, the magnetic moment $\vec{M}$ points along the axis from south to north.
• So the angle between $\vec{M}$ and $\vec{B}$ is the angle given between the axis and the field.
• Here, $\theta = 30^\circ$ is correct. Do not use $90^\circ - 30^\circ = 60^\circ$ unless the problem says “angle with the perpendicular.”

3. Forgetting to Convert Units

Mistake:

Torque is given as $4.5 \times 10^{-2}\ \text{J}$. Since torque has units of N·m, some students mistakenly treat it as energy and try to use work formulas.

How to avoid:

• Torque and energy both have the same SI unit (Joule = N·m), but they are different physical quantities.
• In this formula, $\tau$ is torque, not work. Just plug it in directly — no conversion needed.
• Always check: if the problem says “torque,” use $\tau = MB\sin\theta$.

4. Solving for $M$ Incorrectly

Mistake:

After substituting, students sometimes invert the sine or forget to divide by $\sin\theta$.

How to avoid:

• Write the formula clearly:

• Substitute step-by-step:

Since $\sin 30^\circ = 0.5$:

• Then compute:

• Double-check: The answer should be in A·m² (or J/T). If you get a very small or huge number, re-check the division.

5. Not Stating the Final Answer with Correct Units

Mistake:

Giving $M = 0.36$ without units, or writing wrong units like N·m.

How to avoid:

• Magnetic moment has SI unit A·m² (ampere metre squared) or equivalently J/T (joule per tesla).
• Always write:

• In exams, missing units can cost you marks even if the number is correct.

Quick Summary Checklist

By avoiding these, you’ll solve this problem correctly every time.