Q.A short bar magnet has a magnetic moment of $0.48\ \text{J T}^{-1}$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10\ \text{cm}$ from the centre of the magnet on

Concept: Magnetic field of a bar magnet — the field depends on position (axial vs equatorial) and falls as $1/r^3$.

Step 1 — Axial field formula

For a point on the axis at distance $r$ from the centre:

Direction: away from the north pole (along the axis, from south to north outside the magnet).

Step 2 — Equatorial field formula

For a point on the equatorial line (perpendicular bisector):

Direction: opposite to the magnetic moment (from north to south, parallel to the axis but reversed).

Step 3 — Plug in values

$M = 0.48\ \text{J T}^{-1}$, $r = 0.10\ \text{m}$, $\frac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$. …

The key to solving this lies in understanding that a bar magnet behaves like a magnetic dipole. Its magnetic moment $\mathbf{M}$ is a vector pointing from the south pole to the north pole inside the magnet. The field it produces at any point depends on the orientation of that point relative to the dipole axis.

For a short magnet (length much smaller than the distance $r$), we use the dipole approximation. This is valid here because the distance $10\ \text{cm}$ is large compared to the magnet's length (which is not given but implied to be small). The formulas are exact for a point dipole and excellent approximations for a short bar magnet.

Notice the factor of 2 difference: the axial field is twice the equatorial field at the same distance. This is a direct consequence of the dipole field geometry — field lines are denser along the axis.

Now let's apply these to the given data.

1. Write down the known quantities.

   Magnetic moment $M = 0.48\ \text{J T}^{-1}$ (which is equivalent to $\text{A m}^2$).

   Distance $r = 10\ \text{cm} = 0.10\ \text{m}$.

   The constant $\frac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$ (exactly, by definition of the ampere).

2. Calculate the axial field.

First, $(0.10)^3 = 0.001 = 10^{-3}$.

So $B_{\text{axis}} = 10^{-7} \times \frac{0.96}{10^{-3}} = 10^{-7} \times 0.96 \times 10^{3} = 0.96 \times 10^{-4}\ \text{T}$.

That is $9.6 \times 10^{-5}\ \text{T}$.

Direction: On the axis, the field points away from the north pole and toward the south pole. Since the magnetic moment points from south to north, the axial field is parallel to $\mathbf{M}$ on the side of the north pole, and antiparallel on the south pole side. The problem asks for "direction" — we state it as along the axis, away from the north pole (or equivalently, in the direction of $\mathbf{M}$ if the point is on the north side).

3. Calculate the equatorial field. …

Method: Axial and Equatorial Field Formulas for a Bar Magnet

This problem uses the standard dipole field formulas for a short bar magnet. The magnet is treated as a magnetic dipole, and we apply the two specific cases — axial point and equatorial point.

Step 1 — Recall the formulas

For a short bar magnet of magnetic moment $M$:

• On the axis (end-on position):

• On the equatorial line (broadside-on position):

Where:

• $\frac{\mu_0}{4\pi} = 10^{-7}\ \text{T m A}^{-1}$
• $M = 0.48\ \text{J T}^{-1}$
• $r = 10\ \text{cm} = 0.1\ \text{m}$

Step 2 — Calculate magnitude on the axis

First, $(0.1)^3 = 0.001 = 10^{-3}$

Direction: Along the axis, away from the north pole (i.e., from south to north outside the magnet).

Step 3 — Calculate magnitude on the equatorial line …

Here are the most common mistakes students make with this classic magnetic dipole problem, along with how to avoid each one.

Mistake 1: Using the Wrong Formula for Axis vs. Equator

The Mistake:

Students often mix up the formulas for the magnetic field on the axis ($B_{\text{axis}}$) and the equatorial line ($B_{\text{eq}}$). A common error is using the axis formula for the equator, or forgetting the factor of 2 difference.

The Correct Concept:

The magnetic field due to a short bar magnet (treated as a dipole) is different at these two points.

• On the axis: The field is stronger and points along the direction of the magnetic moment ($\vec{M}$).

• On the equatorial line: The field is half the axial value (at the same distance) and points opposite to the direction of the magnetic moment.

How to Avoid:

• Memorise the ratio: $B_{\text{axis}} = 2 \times B_{\text{eq}}$ (for the same $r$).
• Visualise the dipole: The field lines are denser (stronger) coming out of the north pole (axis) and spread out (weaker) at the sides (equator).

Mistake 2: Forgetting the Unit Conversion for Distance

The Mistake:

The distance is given as $10\ \text{cm}$, but the formula requires metres. A very common error is to plug in $r = 10$ instead of $r = 0.1\ \text{m}$.

The Correct Calculation:

This means $r^3 = (0.1)^3 = 1 \times 10^{-3}\ \text{m}^3$.

How to Avoid:

• Always convert cm to m before plugging into any formula involving SI units (Tesla, J/T, etc.).
• Write the conversion step explicitly in your solution. Do not do it mentally.

Mistake 3: Getting the Direction Wrong

The Mistake:

Students often state the direction incorrectly, especially for the equatorial point. They might say the field points "towards the magnet" without specifying which pole, or they forget the opposite direction on the equator.

The Correct Directions (for a standard bar magnet with North and South poles):

• (a) On the axis: The field points away from the North pole and towards the South pole. In standard notation, if the magnetic moment points from South to North, the field on the axis is along the direction of $\vec{M}$ (from South to North outside the magnet).
• (b) On the equatorial line: The field points parallel to the axis but opposite to the direction of $\vec{M}$ (i.e., from North to South).

How to Avoid:

• Draw a quick sketch. Draw the magnet (N at top, S at bottom). Draw the field lines: they leave N, curve around, and enter S. …