Electrostatic analogy

Every physics student who has just survived the electric-dipole chapter and walks into "Magnetism and Matter" gets a pleasant surprise: half the new chapter is already done. Swap charge $q$ for pole strength $q_m$ , dipole moment $\vec p$ for magnetic moment $\vec m$ , field $\vec E$ for field $\vec B$ , and the constant $\dfrac{1}{4\pi\varepsilon_0}$ for $\dfrac{\mu_0}{4\pi}$ , and almost every electric-dipole formula reappears wearing a magnetic costume — same $1/r^3$ falloff, same factor of two between the axial and equatorial directions, same $\vec\tau=\vec m\times\vec B$ torque law, same $U=-\vec m\cdot\vec B$ energy expression. This is the electrostatic analogy, and it is one of the most exam-efficient ideas in the CBSE Class 12 syllabus — a single correspondence table replaces an entire second derivation you would otherwise have to memorise from scratch. But an analogy that works this well is also dangerous: it tempts you to believe magnetism and electrostatics are the same physics underneath, and NCERT is explicit that they are not. A magnet's north and south poles cannot be separated the way a positive and negative charge can be pulled apart — cut a bar magnet in half and you get two smaller magnets, each with its own N and S pole, never a lone pole in your hand. That single fact is why there is no magnetic Coulomb's law between free poles, and why the magnetic version of Gauss's law gives zero net flux through any closed surface, always. This article builds the correspondence table carefully, walks through where it is exact and where the physics diverges, and reuses the two worked numbers from the shorter Pass-1 version of this section — a $1.5\ \text{A}\,\cdot\,\text{m}^2$ bar magnet's axial field at $0.5\ \text{m}$ , and the universal $B_{eq}=B_{ax}/2$ ratio — while extending them with fresh board- and competitive-level practice.

Key takeaways

The electrostatic analogy is a substitution table. Replace $q\to q_m$ , $\vec p\to\vec m$ , $\vec E\to\vec B$ , $\dfrac{1}{4\pi\varepsilon_0}\to\dfrac{\mu_0}{4\pi}$ and every electric-dipole formula becomes its magnetic-dipole counterpart.
Axial field: $B_{ax}\approx\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ , mirroring $E_{ax}\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ .
Equatorial field: $B_{eq}\approx\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$ , exactly half the axial field at the same distance — in both electrostatics and magnetism.
Torque and energy carry over unchanged in form: $\vec\tau=\vec m\times\vec B$ and $U=-\vec m\cdot\vec B$ , the direct analogues of $\vec\tau=\vec p\times\vec E$ and $U=-\vec p\cdot\vec E$ .
The analogy breaks at the source. Electric dipoles are pairs of separable charges; magnetic dipoles are inseparable N-S pairs (ultimately tiny atomic current loops), so there is no magnetic Coulomb's law and no magnetic Gauss's law with nonzero enclosed "charge."
Both field formulas are short-dipole approximations, valid only when the observation distance $r$ is much larger than the dipole's own length — never apply them close to the magnet.
Key idea: because a magnetic dipole's far-field mathematics has exactly the same 1/r cubed structure as an electric dipole's, the substitution q to q_m, p to m, E to B, and 1/4 pi epsilon_0 to mu_0/4 pi converts any electric-dipole result into its magnetic-dipole twin, even though the two dipoles arise from physically different sources (separable charges versus inseparable current loops).

The core insight

The far fields, torque, and energy of a magnetic dipole obey formulas structurally identical to an electric dipole's under the substitution $q\to q_m$ , $\vec p\to\vec m$ , $\vec E\to\vec B$ and $\dfrac{1}{4\pi\varepsilon_0}\to\dfrac{\mu_0}{4\pi}$ , but the analogy is purely mathematical because magnetic poles, unlike electric charges, can never be isolated.

Where it fits in the syllabus

This analogy sits inside Unit V, "Magnetism and Matter," and functions as NCERT's bridge back to the electric-dipole results of Unit I ("Electric Charges and Fields"). The board syllabus treats the axial and equatorial magnetic-dipole field expressions as qualitative — formulae may be quoted for reference, no derivation is expected — so this section quotes and uses those formulas without deriving them from the Biot-Savart law. Similarly, the torque $\vec\tau=\vec m\times\vec B$ is prescribed only qualitatively. Two things are deliberately kept out here in line with the 2025-26 rationalised syllabus: Earth's magnetism (declination, dip, horizontal component) has been dropped entirely and is not used to motivate this analogy, and Gauss's law for magnetism in vector-calculus form ( $\nabla\cdot\vec B=0$ ) is mentioned only in words, never in that notation — it is developed properly in the sibling node "Magnetism and Gauss's Law."

The physical picture

Picture a short bar magnet lying on a table, its magnetic moment $\vec m$ pointing from its S pole to its N pole. Far away from the magnet — at a distance $r$ much larger than the magnet's own length — the magnetic field it produces has exactly the same geometric shape as the electric field of a short electric dipole placed at the same spot with its moment $\vec p$ pointing the same way. Along the magnet's own axis (an imaginary line through both poles, extended outward), the field is strongest and points along $\vec m$ :

B_{ax} \approx \frac{\mu_0}{4\pi}\cdot\frac{2m}{r^3}

where: $B_{ax}$ is the magnetic field strength (tesla, T) at axial distance $r$ (metres) from the magnet's centre, $m$ is the magnetic moment (A·m $^2$ ), and $\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space. The physical reading is that doubling the distance cuts the axial field to one-eighth, because of the $1/r^3$ dependence — magnetic dipole fields die off far faster than the $1/r^2$ field of a single pole (which does not exist alone anyway). On the perpendicular bisector of the magnet — the equatorial line, at the same distance $r$ from the centre but at right angles to the axis — the field is weaker and points opposite to $\vec m$ :

B_{eq} \approx \frac{\mu_0}{4\pi}\cdot\frac{m}{r^3}

where: all symbols are as above; the missing factor of $2$ compared with the axial case is the single most useful number to memorise in this entire topic, since it lets you jump between the two formulas without re-deriving anything. These are the direct magnetic translations of the electric-dipole fields $E_{ax}\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}$ and $E_{eq}\approx\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ from Unit I — the same shape, scaled by the corresponding constant.

Figure

Two panels showing a bar magnet with its axial field point on the extended centre line and its equatorial field point on the perpendicular bisector, both at the same distance r from the centre, with the axial field arrow drawn twice as long as the equatorial field arrow

The torque a uniform external field $\vec B_{ext}$ exerts on this magnet, trying to align $\vec m$ with $\vec B_{ext}$ , is the direct analogue of the electric-dipole torque:

\vec\tau = \vec m \times \vec B_{ext}, \qquad \tau = mB_{ext}\sin\theta

where: $\theta$ is the angle between $\vec m$ and $\vec B_{ext}$ ; the torque is zero when the magnet is aligned ( $\theta=0$ ) or anti-aligned ( $\theta=180°$ ) and maximum at $\theta=90°$ — precisely the behaviour of an electric dipole in a uniform electric field, $\vec\tau=\vec p\times\vec E$ .

Constructing the correspondence table

Every entry in the table below is obtained by the same three substitutions applied consistently: $q\to q_m$ (charge to pole strength), $\vec p\to\vec m$ (electric to magnetic dipole moment), $\vec E\to\vec B$ (electric to magnetic field), and the constant swap $\dfrac{1}{4\pi\varepsilon_0}\to\dfrac{\mu_0}{4\pi}$ . Electric dipole moment is $\vec p=q(2\vec a)$ , where $2a$ is the separation between $+q$ and $-q$ . The magnetic dipole moment of a bar magnet is defined the same way, $\vec m=q_m(2\vec l)$ , where $2l$ is the effective length between the two poles of pole strength $q_m$ each. Electric field lines start on $+q$ and end on $-q$ ; magnetic field lines run from the N pole outside the magnet to the S pole, then continue inside the magnet from S back to N, forming closed loops — this closed-loop behaviour, with no starting or ending point, is the field-line signature of the fact that magnetic poles cannot be isolated. Energy of an electric dipole in a uniform field is $U=-\vec p\cdot\vec E$ , minimum (most stable) when $\vec p$ is aligned with $\vec E$ . The magnetic analogue is $U=-\vec m\cdot\vec B$ , minimum when the magnet's moment aligns with the external field — the physical reason a compass needle settles pointing along the local field rather than across it or against it.

Derivation AUsing the analogy to write down the magnetic dipole's axial field

The goal is to show, step by step, exactly how the substitution rule converts the known electric-dipole axial field into the magnetic-dipole axial field quoted above — not by deriving either field from Biot-Savart or Coulomb's law (both are beyond this qualitative topic), but by applying the same three-part substitution consistently to a formula the student already trusts from Unit I, and confirming the result matches the reference formula. This matters because the same mechanical substitution then works for every other electric-dipole formula in the syllabus, so mastering it once here saves re-deriving the whole magnetism chapter from scratch.

Step 1. Start from the trusted electric-dipole axial field, valid for $r\gg a$ where $2a$ is the charge separation: $E_{ax}=\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2p}{r^3}$ , where $p=q(2a)$ is the electric dipole moment.

Step 2. Apply the constant substitution first: replace $\dfrac{1}{4\pi\varepsilon_0}$ , the electrostatic constant, with $\dfrac{\mu_0}{4\pi}$ , the corresponding magnetic constant. The formula becomes $E_{ax}\to\dfrac{\mu_0}{4\pi}\cdot\dfrac{2p}{r^3}$ — still written with the electric symbol $p$ and $E$ for now, only the constant has changed.

Step 3. Apply the quantity substitutions: replace the field symbol $E\to B$ and the dipole moment $p\to m$ , since $m=q_m(2l)$ plays the exact same structural role as $p=q(2a)$ — both are "pole (or charge) strength times separation." The formula becomes $B_{ax}=\dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^3}$ .

Step 4. Compare against the reference formula quoted in the syllabus, $B\approx\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}$ . They match exactly. The identical procedure applied to the equatorial formula $E_{eq}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}$ produces $B_{eq}=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^3}$ , again matching the reference formula — confirming the substitution rule is self-consistent across both the axial and equatorial cases.

Worked examples

Board level - axial field of a bar magnet

A short bar magnet has magnetic moment $m=1.5\ \text{A}\,\cdot\,\text{m}^2$ . Find the magnetic field on its axis at a distance $r=0.5\ \text{m}$ from its centre. Take $\mu_0=4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ .

Step 1. Write down the axial formula obtained by the analogy: $B_{ax}\approx\dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{r^3}$ , and note that $\dfrac{\mu_0}{4\pi}=10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ once the $4\pi$ cancels — a simplification worth memorising for every numerical in this topic.

Step 2. Substitute the given numbers: $B_{ax}\approx 10^{-7}\times\dfrac{2\times1.5}{(0.5)^3}=10^{-7}\times\dfrac{3.0}{0.125}$ .

Step 3. Evaluate the fraction and finish: $B_{ax}\approx 10^{-7}\times 24 = 2.4\times10^{-6}\ \text{T}$ .

The axial field is $\boldsymbol{2.4\times10^{-6}\ \text{T}}$ , obtained entirely by substituting into the electric-dipole axial formula's magnetic counterpart — no fresh derivation was needed.

Competitive level - equatorial field and torque together

Using the same magnet ( $m=1.5\ \text{A}\,\cdot\,\text{m}^2$ ), a point P lies on the equatorial line at the same distance $r=0.5\ \text{m}$ from the centre. Find (a) the field at P, and (b) the torque on this magnet if it is placed with its moment at $30°$ to a uniform external field of magnitude $0.02\ \text{T}$ .

Step 1. Use the universal ratio proved in this topic, $B_{eq}=B_{ax}/2$ , instead of recomputing from scratch: since $B_{ax}=2.4\times10^{-6}\ \text{T}$ from the board-level example, $B_{eq}=1.2\times10^{-6}\ \text{T}$ immediately.

Step 2. Check this independently from the formula: $B_{eq}\approx 10^{-7}\times\dfrac{1.5}{0.125}=10^{-7}\times 12=1.2\times10^{-6}\ \text{T}$ — matches Step 1, confirming the shortcut is reliable and saves time under exam pressure.

Step 3. For the torque, use $\tau=mB_{ext}\sin\theta$ with $m=1.5\ \text{A}\,\cdot\,\text{m}^2$ , $B_{ext}=0.02\ \text{T}$ , $\theta=30°$ , so $\sin30°=0.5$ : $\tau=1.5\times0.02\times0.5=0.015\ \text{N}\,\cdot\,\text{m}$ .

The equatorial field is $\boldsymbol{1.2\times10^{-6}\ \text{T}}$ and the torque is $\boldsymbol{0.015\ \text{N}\,\cdot\,\text{m}}$ , tending to rotate the magnet so its moment aligns with the external field. Timing note: in a competitive-exam multi-part question like this, recognising the $B_{eq}=B_{ax}/2$ shortcut (Step 1) and reusing an already-computed number is worth roughly 20-30 seconds saved over recomputing from the raw formula — small on one question, significant across a full paper of dipole numericals.

Common traps

Treating the analogy as proof that isolated magnetic poles exist "in some sense" — they do not; a magnet cut in half always yields two complete dipoles, never a lone pole, and this is a genuine physical fact, not a limitation of measurement.
Forgetting the constant swap and mixing $\dfrac{1}{4\pi\varepsilon_0}$ with $\mu_0$ -based formulas, which gives an answer with the right structure but a wrong numerical value by many orders of magnitude.
Assuming Gauss's law transfers unchanged: the electric version gives nonzero flux through a surface enclosing net charge, but the magnetic version gives exactly zero net flux through any closed surface, because there is no magnetic "charge" to enclose — this equality with zero, not a mere similarity, is the content of Gauss's law for magnetism.
Using the axial or equatorial formulas at distances comparable to the magnet's own physical length — both formulas are short-dipole approximations valid only for $r\gg l$ , and applying them close to the magnet gives numbers with no physical meaning.
Forgetting the missing factor of two between axial and equatorial fields and writing $B_{eq}=B_{ax}$ by mistake, a very common one-mark slip under time pressure.

You should be able to…

Reproduce the correspondence table between electric-dipole and magnetic-dipole quantities: $q\leftrightarrow q_m$ , $\vec p\leftrightarrow\vec m$ , $\vec E\leftrightarrow\vec B$ , $\dfrac{1}{4\pi\varepsilon_0}\leftrightarrow\dfrac{\mu_0}{4\pi}$ .
Use the analogy to write down a magnetic-dipole formula (axial field, equatorial field, torque, energy) from its electric-dipole counterpart, and vice versa.
State and justify the key physical difference: electric charges are separable, magnetic poles are not — and connect this to why magnetic field lines form closed loops while electric field lines do not.
Explain, in words only, why the analogy does not extend to Gauss's law, without writing the vector-calculus form $\nabla\cdot\vec B=0$ .
Compute axial and equatorial fields, and torque and energy, for a magnetic dipole using the analogy, including recognising the $B_{eq}=B_{ax}/2$ shortcut in numericals.
Recall that both dipole-field formulas are short-dipole approximations valid only for $r$ much greater than the dipole's own length.

Ask the tutor: Want to practise converting a few more electric-dipole results — like the electric dipole's field at a general angle — into their magnetic counterparts using this same substitution rule?

If the formulas are identical in shape, why does my teacher keep saying magnetism and electrostatics are different subjects?

Because "identical shape" only describes the far-field mathematics, not the underlying physics. An electric dipole is two real, separable charges sitting a fixed distance apart — you could in principle carry the $+q$ to Delhi and the $-q$ to Mumbai. A magnetic dipole cannot be separated that way; cut a bar magnet anywhere and you get two smaller complete magnets, each with its own N and S pole. At the microscopic level, magnetism comes from circulating currents (electron motion), not from any magnetic charge sitting at a point. The analogy is a computational convenience for the far field, not a claim that the two phenomena share a common physical cause.

Does the analogy also work for the near field, close to the magnet or dipole, not just far away?

No — both formulas quoted here, axial and equatorial, are short-dipole (far-field) approximations, valid only when $r$ is much larger than the dipole's own length. Close to a real bar magnet, the field lines bulge out near the poles and the simple $1/r^3$ formulas no longer describe the field accurately; the true near-field pattern depends on the magnet's actual shape and pole distribution. The syllabus only asks you to use and quote the far-field formulas, so for CBSE purposes always assume the " $r$ is large" condition is satisfied unless a question tells you otherwise.

Since the analogy fails for Gauss's law, does that mean it also fails for the torque and energy formulas?

No — torque and energy carry over cleanly because they depend only on how a dipole (electric or magnetic) responds to being placed in a uniform external field, and that response is governed by the same vector geometry, $\vec\tau=\vec m\times\vec B$ paralleling $\vec\tau=\vec p\times\vec E$ , and $U=-\vec m\cdot\vec B$ paralleling $U=-\vec p\cdot\vec E$ , in both cases. Gauss's law fails to carry over for a different reason entirely: it depends on whether a "charge" can be enclosed inside a surface, and since isolated magnetic pole-charge does not exist, the enclosed magnetic charge is always zero, giving zero flux always — a statement about sources, not about how a dipole sits in an external field.

Is this analogy actually asked as a standalone question in board exams, or is it just a study trick?

Both. NCERT itself presents the correspondence as a table the student should be able to reproduce, and board papers do ask direct one- or two-mark questions like "state the analogy between the electrostatic dipole and the magnetic dipole" or "using the analogy, write the expression for the magnetic field on the axis of a magnetic dipole." So it is examinable content in its own right, not merely a memory aid — though it is also genuinely useful as a memory aid whenever you blank on which magnetism formula goes where.