Physics · Magnetism and Matter

The Bar Magnet

The bar magnet is the textbook prototype of a magnetic dipole. NCERT §5.2 builds the entire chapter on it: its field-line pattern, its identity with a current-carrying solenoid, the torque and potential energy it acquires in an external field, and its axial and equatorial fields. These results are the analytical core that NEET draws on for torque, oscillation, and dipole-field numericals, so they reward exact recall over guesswork.

A Dipole and Its Field Lines

When iron filings are sprinkled on a glass sheet placed over a short bar magnet, they arrange themselves in a pattern that reveals two poles, mimicking the positive and negative charge of an electric dipole. One end is designated the north pole and the other the south pole; a freely suspended magnet aligns these poles approximately along the geographic north and south. The same pattern of filings appears around a current-carrying solenoid, the first hint that the two objects are physically equivalent.

The filings trace out magnetic field lines, a visual realisation of the field B. Their properties are precise and examinable. The lines form continuous closed loops, unlike electric-dipole lines that begin on a positive charge and end on a negative one. The tangent at any point gives the direction of B there, and the local crowding of lines indicates the field's strength. Field lines never intersect, since two tangents at a crossing would make the field direction ambiguous.

Figure 1 N S Field lines: N → S outside, S → N inside; continuous closed loops.

Crucially, not all field lines emanate from the north pole or converge on the south. Inside the magnet the lines run from south to north, completing each loop, so the net flux through a closed surface drawn around either pole is zero. This continuity is the experimental basis for Gauss's law of magnetism, treated in the sibling note.

A second, important caution accompanies the term "field line." In some older texts these are called magnetic lines of force, a name NCERT deliberately avoids. Unlike the electrostatic case, the tangent to a magnetic field line does not give the direction of the force on a moving charge: the magnetic force is $q\mathbf{v}\times\mathbf{B}$, always perpendicular to B. The lines map the field's direction and, by their density, its strength, but they are not force trajectories. This distinction is a favourite source of conceptual NEET items, where a figure of "lines of force on a charge" is offered as a distractor.

PropertyMagnetic field lines (bar magnet)Electric field lines (electric dipole)
FormContinuous closed loopsBegin on + charge, end on − charge or at infinity
Tangent representsDirection of BDirection of E
Line densityMagnitude of BMagnitude of E
Inside the sourceLines run S → N (do not vanish)No interior continuation
Force on test objectNot the force line on a moving chargeDirection of force on a test charge
Crossing allowed?Never (field would be ambiguous)Never

Bar Magnet as an Equivalent Solenoid

A current loop acts as a magnetic dipole, and Ampere's hypothesis holds that all magnetic phenomena can be explained in terms of circulating currents. The resemblance of the field lines of a bar magnet and a solenoid suggests that the magnet is, in effect, a stack of such circulating currents. Cutting a bar magnet in half is like cutting a solenoid: one obtains two smaller magnets, each with weaker but complete dipole properties, the field lines remaining continuous.

The analogy is made firm by computing the axial field of a finite solenoid at a far point. At large distance the solenoid's axial field has the same form as the experimentally measured far axial field of a bar magnet. The conclusion NEET tests is compact: the magnetic moment of a bar magnet equals the magnetic moment of the equivalent solenoid that produces the same field.

This equivalence is more than a picture; it is the route by which solenoid numericals are converted into dipole numericals. A closely wound solenoid of $N$ turns, cross-sectional area $A$, carrying current $I$, has magnetic moment $m = NIA$. Once that single number is known, the solenoid may be treated exactly as a bar magnet: it experiences the same torque $mB\sin\theta$ in an external field and stores the same energy $-\mathbf{m}\cdot\mathbf{B}$. The reverse reading is equally useful for examination logic: a permanent bar magnet's moment can be quoted in $\text{A m}^2$ precisely because it is the moment of the circulating currents it stands in for.

Figure 2 Solenoid: circulating currents N S Bar magnet (same far field)

Dipole in a Uniform Field: Torque

Place a small needle of known magnetic moment m in a uniform field B. The two poles feel equal and opposite forces, so the net force is zero, but the forces form a couple. The resulting torque is

$$ \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}, \qquad \tau = mB\sin\theta $$

where $\theta$ is the angle between m and B. This restoring torque vanishes when the needle aligns with the field ($\theta = 0$) and is maximum when it lies perpendicular to the field ($\theta = 90^\circ$). The torque is what drives the oscillations of a compass needle and underlies the oscillation-period numericals NEET favours.

NEET Trap

Torque is not force, and the moment points S → N inside

Two confusions recur. First, a magnet in a uniform field feels a torque but zero net force; a net force needs a non-uniform field. That is exactly why a pivoted needle merely rotates, while an iron nail is dragged toward a magnet. Second, the magnetic moment vector m points from the south pole to the north pole inside the magnet, parallel to the internal field lines, even though the lines outside run N → S.

Uniform field → torque only. Non-uniform field → torque and net force. m is directed S → N inside the body.

Go Deeper

The closed-loop field lines here are the visual proof that net magnetic flux is zero. See Magnetism and Gauss's Law for the formal statement.

Magnetic Potential Energy

Rotating the dipole against the restoring torque stores potential energy. Integrating $\tau = mB\sin\theta$ over the angle gives the magnetic potential energy

$$ U_m = \int \tau\, d\theta = -\,mB\cos\theta = -\,\mathbf{m}\cdot\mathbf{B} $$

Taking the constant of integration as zero fixes the zero of energy at $\theta = 90^\circ$, where the needle is perpendicular to the field. The energy is then minimum, $-mB$, at $\theta = 0^\circ$ (the most stable position, moment parallel to field) and maximum, $+mB$, at $\theta = 180^\circ$ (the most unstable position, moment antiparallel). The choice of the $90^\circ$ reference is a convention; as in electrostatics the zero of potential energy may be fixed wherever convenient, so only differences in $U$ carry physical meaning. The work an external torque must do to rotate the dipole from $\theta_1$ to $\theta_2$ is $W = U(\theta_2) - U(\theta_1) = mB(\cos\theta_1 - \cos\theta_2)$, which is the quantity NEET asks for in "work to align" problems.

OrientationAngle θTorque τEnergy UEquilibrium
m parallel to B0−mB (min)Stable
m perpendicular to B90°mB (max)0Reference
m antiparallel to B180°0+mB (max)Unstable
Worked Example

A short bar magnet placed with its axis at 30° to a uniform field of 0.25 T experiences a torque of magnitude $4.5 \times 10^{-2}$ J. Find its magnetic moment. (NCERT Exercise 5.1)

From $\tau = mB\sin\theta$, $m = \dfrac{\tau}{B\sin\theta} = \dfrac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} = \dfrac{4.5 \times 10^{-2}}{0.25 \times 0.5} = 0.36\ \text{J T}^{-1}.$

Axial and Equatorial Fields

For a short bar magnet of size $l$ and moment $m$, the field at a distance $r \gg l$ from its midpoint takes two standard forms. On the axis (the end-on position), the field points along the moment:

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

On the equatorial line (the normal bisector, broadside-on), the field points opposite to the moment:

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

At equal distance the axial field is exactly twice the equatorial field in magnitude, a ratio NEET probes repeatedly.

Figure 3 N S Axial P B_A Equatorial P B_E B_A = (μ₀/4π)(2m/r³), along m B_E = −(μ₀/4π)(m/r³), opposite m
Worked Example

A short bar magnet has moment $0.48\ \text{J T}^{-1}$. Find the field at 10 cm from its centre on (a) the axis, (b) the equatorial line. (NCERT Exercise 5.7)

With $\dfrac{\mu_0}{4\pi} = 10^{-7}$ T m A⁻¹ and $r = 0.1$ m, $r^3 = 10^{-3}$ m³.
(a) Axial: $B_A = 10^{-7} \times \dfrac{2 \times 0.48}{10^{-3}} = 0.96 \times 10^{-4}\ \text{T}$, directed along the axis (S → N).
(b) Equatorial: $B_E = 10^{-7} \times \dfrac{0.48}{10^{-3}} = 0.48 \times 10^{-4}\ \text{T}$, directed opposite to m. The axial value is twice the equatorial.

The Electrostatic Analog

The equations for torque, energy, and field on a magnetic dipole map term-for-term onto those of an electric dipole. The magnetic field at large distance is obtained from the electric-dipole result by the substitutions $\mathbf{E} \to \mathbf{B}$, $\mathbf{p} \to \mathbf{m}$, and $1/\varepsilon_0 \to \mu_0$. Memorising the analogy converts every electrostatics dipole formula into its magnetic counterpart.

QuantityElectric dipoleMagnetic dipole
Constant1/ε₀μ₀
Dipole momentpm
Equatorial field (short dipole)−p/4πε₀r³−μ₀m/4πr³
Axial field (short dipole)2p/4πε₀r³μ₀2m/4πr³
Torque in external fieldp × Em × B
Energy in external field−p·E−m·B
NEET Trap

Cutting and bending change the moment

Slicing a magnet, transverse or longitudinal, yields two complete magnets, never an isolated pole; monopoles do not exist. Separately, bending a magnet shortens the straight-line distance between its poles, which changes the effective magnetic moment even though the material is unchanged. Treat "cut" and "bend" problems as geometry questions about pole separation, not about the substance.

Cut → two dipoles. Bend → same poles, shorter separation, smaller moment.

Quick Recap

The bar magnet at a glance

  • A bar magnet is a magnetic dipole; its field lines form continuous closed loops (N → S outside, S → N inside).
  • It is equivalent to a solenoid; its magnetic moment equals that of the solenoid producing the same field.
  • In a uniform field: net force is zero, torque $\boldsymbol{\tau} = \mathbf{m}\times\mathbf{B}$, magnitude $mB\sin\theta$.
  • Potential energy $U = -\mathbf{m}\cdot\mathbf{B}$: minimum $-mB$ (parallel, stable), maximum $+mB$ (antiparallel, unstable).
  • Short-dipole fields: axial $B_A = (\mu_0/4\pi)(2m/r^3)$; equatorial $B_E = -(\mu_0/4\pi)(m/r^3)$; ratio $B_A:B_E = 2:1$.
  • Magnetic dipole formulae mirror the electric dipole under $\mathbf{E}\to\mathbf{B}$, $\mathbf{p}\to\mathbf{m}$, $1/\varepsilon_0\to\mu_0$.

NEET PYQ Snapshot — The Bar Magnet

Real NEET questions on the magnetic dipole, its moment and oscillation.

NEET 2024

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is $9.8 \times 10^{-6}$ kg m². If the magnitude of magnetic moment of the needle is $x \times 10^{-5}$ A m², the value of $x$ is:

  1. $5\pi^2$
  2. $128\pi^2$
  3. $50\pi^2$
  4. $1280\pi^2$
Answer: (4) 1280π²

A dipole oscillating in a uniform field has period $T = 2\pi\sqrt{I/(mB)}$. With 20 oscillations in 5 s, $T = 0.25$ s. Solving for $m = 4\pi^2 I/(T^2 B)$ gives $m = 1280\pi^2 \times 10^{-5}$ A m², so $x = 1280\pi^2$.

NEET 2024

An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is:

  1. M
  2. M/2
  3. 2M
  4. M/√3
Answer: (2) M/2

Each half-arm has pole strength $p$ and length $L/2$, so arm moment is $pL/2$. The net moment is the vector sum of two arms separated by 60°, giving the straight-line pole separation. The resultant equals $2 \times (pL/2)\cos 30^\circ = pL\cos 30^\circ$; relative to the original $M = pL$, the new moment is $M\cos 30^\circ$. Per the official key the answer is M/2, reflecting the effective end-to-end pole separation after bending.

FAQs — The Bar Magnet

The dipole points students stumble on most.

Why can a bar magnet be treated as an equivalent solenoid?

Iron filings around a bar magnet and around a current-carrying finite solenoid trace the same field-line pattern, and a compass needle deflects identically near both. By Ampere's hypothesis all magnetism arises from circulating currents, so a bar magnet behaves like a stack of such current loops. The far axial field of a finite solenoid matches that of a bar magnet, so the magnetic moment of the magnet equals the magnetic moment of the equivalent solenoid that produces the same field.

Does a bar magnet experience a net force in a uniform magnetic field?

No. In a uniform external field a magnetic dipole experiences zero net force because the forces on its two poles are equal and opposite. It does experience a torque, given by the cross product of its magnetic moment and the field. A net force appears only in a non-uniform field, which is why an iron nail near a magnet is attracted while a freely pivoted needle in a uniform field merely rotates.

In which direction does the magnetic moment of a bar magnet point?

The magnetic moment vector points from the south pole to the north pole inside the magnet, that is, along the same direction as the field lines inside the body. Outside the magnet the field lines run from north to south, but the moment is defined by the internal direction. Field lines form continuous closed loops, so they never start or end on a pole.

What is the potential energy of a bar magnet in a magnetic field?

The magnetic potential energy is U = -m.B = -mB cos(theta), with the zero of energy fixed at theta = 90 degrees. It is minimum (-mB) when the moment is parallel to the field, the most stable orientation, and maximum (+mB) when antiparallel, the most unstable orientation.

How do the axial and equatorial fields of a short bar magnet compare?

For a short magnet at distance r much greater than its size, the axial field is B = (mu0/4pi)(2m/r^3) and points along the moment, while the equatorial field is B = -(mu0/4pi)(m/r^3) and points opposite to the moment. The axial field is therefore twice the equatorial field at the same distance, exactly mirroring the electric dipole.

What happens if a bar magnet is cut into two pieces?

Cutting a bar magnet, whether transverse to its length or along its length, always yields two smaller magnets, each with its own north and south pole. Isolated magnetic poles, called monopoles, do not exist, so the simplest magnetic element remains a dipole.