The Electrostatic Analog

The Translation Dictionary

Here's one of the most elegant shortcuts in Class 12 physics. The far field of a bar magnet has exactly the same mathematical form as the field of an electric dipole. So every result you derived in Chapter 1 can be translated into magnetism with three replacements:

$$\vec{E} \to \vec{B}, \qquad \vec{p} \to \vec{m}, \qquad \frac{1}{4\pi\varepsilon_0} \to \frac{\mu_0}{4\pi}$$

The complete dipole analogy:

Quantity	Electrostatics	Magnetism
Constant in dipole-field formula	$\dfrac{1}{4\pi\varepsilon_0}$	$\dfrac{\mu_0}{4\pi}$
Dipole moment	$\vec{p}$	$\vec{m}$
Equatorial field (short dipole)	$-\dfrac{\vec{p}}{4\pi\varepsilon_0 r^3}$	$-\dfrac{\mu_0 \vec{m}}{4\pi r^3}$
Axial field (short dipole)	$\dfrac{2\vec{p}}{4\pi\varepsilon_0 r^3}$	$\dfrac{\mu_0\, 2\vec{m}}{4\pi r^3}$
Torque in external field	$\vec{p} \times \vec{E}$	$\vec{m} \times \vec{B}$
Energy in external field	$-\vec{p} \cdot \vec{E}$	$-\vec{m} \cdot \vec{B}$

[JEE Tip] Don't memorise magnetism formulas separately — memorise the dictionary. One table, two chapters.

Axial and Equatorial Fields of a Bar Magnet

For a bar magnet of size $l$ and moment $\vec{m}$, at distance $r \gg l$ from its centre:

Equatorial field (on the perpendicular bisector):

$$\vec{B}_E = -\frac{\mu_0 \vec{m}}{4\pi r^3}$$

Axial field (on the axis):

$$\vec{B}_A = \frac{\mu_0\, 2\vec{m}}{4\pi r^3}$$

Read the signs carefully — they carry the physics:

• On the axis, $\vec{B}_A$ is parallel to $\vec{m}$. On the N-side axial point it points away from the N pole; on the S-side axial point it points towards the S pole, but in both cases it is parallel to $\vec{m}$.
• On the equator, $\vec{B}_E$ is antiparallel to $\vec{m}$ — the minus sign! A compass on the perpendicular bisector points opposite to the magnet's moment.
• At the same distance, the axial field magnitude is twice the equatorial field magnitude: $|\vec{B}_A| = 2|\vec{B}_E|$.
• Both fall as $1/r^3$ — the universal dipole signature.

[NEET Important] If a short bar magnet's axial field at distance $r$ equals its equatorial field at distance $r'$, then setting $\dfrac{2m}{r^3} = \dfrac{m}{r'^3}$ gives $r = 2^{1/3}r'$. This is a common exam-pattern comparison.

Two Dipoles Together

Place a dipole Q near a fixed dipole P. Q sits in P's field $\vec{B}_P$, so its energy is $U = -\vec{m}_Q \cdot \vec{B}_P$. The rules:

• Equilibrium is stable when $\vec{m}_Q$ is parallel to $\vec{B}_P$, unstable when antiparallel.
• On P's axis, $\vec{B}_P$ points along $\vec{m}_P$ (strength $\propto 2/r^3$).
• On P's normal bisector, $\vec{B}_P$ points opposite to $\vec{m}_P$ (strength $\propto 1/r^3$).

So at the same distance $r$:

Position of Q	Orientation of $\vec{m}_Q$	Equilibrium	Energy
On axis	parallel to $\vec{m}_P$	Stable	$-\dfrac{\mu_0 2 m_P m_Q}{4\pi r^3}$ (lowest!)
On axis	antiparallel to $\vec{m}_P$	Unstable	$+\dfrac{\mu_0 2 m_P m_Q}{4\pi r^3}$
On bisector	antiparallel to $\vec{m}_P$	Stable	$-\dfrac{\mu_0 m_P m_Q}{4\pi r^3}$
On bisector	parallel to $\vec{m}_P$	Unstable	$+\dfrac{\mu_0 m_P m_Q}{4\pi r^3}$
Either place	perpendicular to $\vec{B}_P$	Not in equilibrium (torque acts)	0

The lowest-energy configuration of all: Q on the axis with both moments aligned — because the axial field is twice as strong.

[JEE Tip] At a general point at angle $\theta$ from the axis (distance $r$), the magnitude is $B = \dfrac{\mu_0 m}{4\pi r^3}\sqrt{1 + 3\cos^2\theta}$ — it interpolates between $2m$-type (axis, $\theta = 0$) and $m$-type (equator, $\theta = 90^{\circ}$). Beyond board level, but standard in JEE.

Solved Examples

Example 1: Axial field of a short magnet

A short bar magnet has magnetic moment 0.48 J/T. Find the magnetic field at a distance of 10 cm from its centre on the axis.

Solution:

1. Formula: $B_A = \dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ with $\dfrac{\mu_0}{4\pi} = 10^{-7}$ T m/A.
2. Substitute: $r = 0.1$ m, $r^3 = 10^{-3}$ m$^3$: $B_A = 10^{-7} \times \dfrac{2 \times 0.48}{10^{-3}}$.
3. Calculate: $B_A = 10^{-7} \times 960 = 0.96 \times 10^{-4}$ T.
4. Answer: $0.96 \times 10^{-4}$ T (0.96 gauss), directed along the moment (from S to N).

Example 2: Equatorial field of the same magnet

For the same magnet, find the field at 10 cm on the equatorial (perpendicular bisector) line.

Solution:

1. Magnitude formula: $|\vec{B}_E| = \dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$ — half the axial value at the same $r$. Vectorially, $\vec{B}_E$ is opposite to $\vec{m}$.
2. $|\vec{B}_E| = 10^{-7} \times \dfrac{0.48}{10^{-3}} = 0.48 \times 10^{-4}$ T.
3. Answer: $0.48 \times 10^{-4}$ T (0.48 gauss), directed opposite to the moment $\vec{m}$.

Example 3: Where do axial and equatorial fields match?

At what distance on the axis does a short magnet produce the same field magnitude as it produces at 10 cm on the equator?

Solution:

1. Set equal: $\dfrac{2m}{r_A^3} = \dfrac{m}{r_E^3}$ with $r_E = 0.1$ m.
2. $r_A^3 = 2 r_E^3$, so $r_A = 2^{1/3} \times 0.1 = 1.26 \times 0.1$ m.
3. Answer: $r_A \approx 12.6$ cm.

Example 4: Compass direction on the bisector

A compass is placed on the perpendicular bisector of a short bar magnet, far from it. Which way does its north tip point?

Solution:

1. On the equatorial line, $\vec{B}_E = -\dfrac{\mu_0\vec{m}}{4\pi r^3}$ — antiparallel to $\vec{m}$.
2. The compass aligns along the local field, so its north tip points opposite to the magnet's moment (i.e. in the N-to-S direction of the magnet).
3. On the axis it would instead point along $\vec{m}$.

Example 5: Stable configurations of two needles

Dipole Q is placed at distance $r$ from dipole P, either on P's axis or on its normal bisector, with $\vec{m}_Q$ parallel or antiparallel to $\vec{m}_P$. Which configurations are in stable equilibrium?

Solution:

1. Rule: stable when $\vec{m}_Q \parallel \vec{B}_P$ at Q's location.
2. On the axis: $\vec{B}_P \parallel \vec{m}_P$. Stable when $\vec{m}_Q$ is parallel to $\vec{m}_P$.
3. On the bisector: $\vec{B}_P$ is antiparallel to $\vec{m}_P$. Stable when $\vec{m}_Q$ is antiparallel to $\vec{m}_P$.
4. The reversed orientations in each position are unstable; perpendicular orientations are not equilibria at all (torque acts).

Example 6: The lowest-energy configuration

Among all the stable configurations of Example 5 (same $r$), which has the lowest potential energy?

Solution:

1. $U = -m_Q B_P$, so the lowest energy needs the largest $B_P$.
2. The axial field ($\propto 2/r^3$) is twice the equatorial field ($\propto 1/r^3$).
3. Answer: Q on P's axis with moments parallel: $U = -\dfrac{\mu_0 2 m_P m_Q}{4\pi r^3}$ — twice as negative as the bisector case.

Example 7: Interaction energy numerical

Two short magnets, each of moment 1.0 A m$^2$, are coaxial with their moments aligned, centres 1.0 m apart. Find their interaction energy.

Solution:

1. Field of magnet 1 at magnet 2 (axial): $B = 10^{-7} \times \dfrac{2 \times 1.0}{1.0^3} = 2 \times 10^{-7}$ T.
2. Energy: $U = -m_2 B = -1.0 \times 2 \times 10^{-7}$.
3. Answer: $U = -2 \times 10^{-7}$ J. Negative — the aligned coaxial pair is bound (stable).

Example 8: Translating an electrostatics result

The electric field of a short dipole at a general point is $E = \dfrac{p}{4\pi\varepsilon_0 r^3}\sqrt{1+3\cos^2\theta}$. Write the magnetic counterpart.

Solution:

1. Apply the dictionary: $p \to m$, $1/4\pi\varepsilon_0 \to \mu_0/4\pi$, $E \to B$.
2. Answer: $B = \dfrac{\mu_0 m}{4\pi r^3}\sqrt{1+3\cos^2\theta}$.
3. Check the limits: $\theta = 0$ gives $\dfrac{\mu_0 2m}{4\pi r^3}$ (axial); $\theta = 90^{\circ}$ gives $\dfrac{\mu_0 m}{4\pi r^3}$ (equatorial). The dictionary works perfectly.

Example 9: Field at a general point (JEE pattern)

A short magnet has moment 1.0 A m$^2$. Find the field magnitude at 10 cm from its centre, at 60 degrees from the axis.

Solution:

1. Formula: $B = \dfrac{\mu_0 m}{4\pi r^3}\sqrt{1+3\cos^2\theta}$.
2. $\cos 60^{\circ} = 0.5$, so $1 + 3(0.25) = 1.75$, $\sqrt{1.75} \approx 1.32$.
3. $\dfrac{\mu_0 m}{4\pi r^3} = 10^{-7} \times \dfrac{1.0}{10^{-3}} = 10^{-4}$ T.
4. Answer: $B \approx 1.32 \times 10^{-4}$ T.

Example 10: Same field, two distances

The equatorial field of a magnet at 5 cm is $B_0$. At what axial distance is the field also $B_0$?

Solution:

1. $B_0 = \dfrac{\mu_0 m}{4\pi (0.05)^3}$ and we need $\dfrac{\mu_0 2m}{4\pi r^3} = B_0$.
2. Dividing: $\dfrac{2 (0.05)^3}{r^3} = 1$, so $r = 2^{1/3} \times 5$ cm $\approx 6.3$ cm.
3. Takeaway: the factor 2 always shows up as a $2^{1/3} \approx 1.26$ stretch in distance.