JEE Mains Physics · Magnetism and Matter

Bar Magnets and Magnetic Dipoles

A bar magnet, a current loop and an orbiting electron are all magnetic dipoles: each has a moment M, makes a field that falls as 1/r³, and in a uniform field feels a torque MB sin θ with potential energy −MB cos θ.

All Magnetism and Matter notes JEE Mains Physics strategy

Why this matters

Fourteen PYQs, six of them asking for a number, and two from 2026. Eight put a magnet or a coil in a uniform field and ask for the torque, the potential energy or the work to turn it; six are about the dipole itself: its moment, its field and potential, or the absence of single poles. The arithmetic is short, so the marks go on the angle and the sign.

Concept 1 of 2: Magnetic dipole moment and the field of a bar magnet

A bar magnet behaves like two poles of strength m a distance 2l apart, so its moment is m × 2l, pointing from S to N. A coil of current is the same kind of object with moment NIA. Far away, the field on the axis is twice the field on the equator at the same distance, and both fall as 1/r³. There are no single magnetic poles, so every field line closes on itself.

Definition

Bar magnet: $M = m(2l)$ , directed from S to N; unit $A\,m^{2}$ , the same as $J\,T^{-1}$ .
Current loop: $M = NIA$ , along the normal to its plane (right-hand rule). An orbiting electron: $\vec{\mu} = -\dfrac{e}{2m_e}\vec{L}$ , opposite to $\vec{L}$ because its charge is negative.
No monopoles: field lines are closed loops, and the net magnetic flux through any closed surface is zero.
On the axis: $B = \dfrac{\mu_0}{4\pi}\dfrac{2Mr}{(r^{2} - l^{2})^{2}}$ , which becomes $\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^{3}}$ for a short magnet; it points along M.
On the equator: $B = \dfrac{\mu_0}{4\pi}\dfrac{M}{(r^{2} + l^{2})^{3/2}}$ , which becomes $\dfrac{\mu_0}{4\pi}\dfrac{M}{r^{3}}$ ; it points opposite to M.
Magnetic potential on the axis: $V = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^{2}}$ ; it is zero on the equator.
Fields of two magnets add as vectors; two fields at right angles combine as $\sqrt{B_1^{2} + B_2^{2}}$ .
Bending a magnet keeps its pole strength but brings the poles closer. Bent into a semicircle: $M' = 2M/\pi$ . Bent at its middle into an L: $M' = M/\sqrt{2}$ .
Error in a field: add the relative errors, each times its power; $(r^{2} + l^{2})^{3/2}$ contributes $\tfrac{3}{2}$ times the relative error of $r^{2} + l^{2}$ .

Short magnet: axial field, equatorial field, axial potential

B_{axial} = \frac{\mu_0}{4\pi}\frac{2M}{r^{3}} \qquad B_{eq} = \frac{\mu_0}{4\pi}\frac{M}{r^{3}} \qquad V_{axis} = \frac{\mu_0}{4\pi}\frac{M}{r^{2}}

Worked example

A short bar magnet has pole strength

10\ A\,m

and magnetic length 4 cm. Find its moment, then the field at 20 cm from its centre on the axis and on the equator, and the magnetic potential at the axial point.

(\mu_0/4\pi = 10^{-7}\ T\,m\,A^{-1})

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q111Moderate

Example 1 · Magnetism and Matter · Bar Magnets and Magnetic Dipoles

A straight magnetic strip has a magnetic moment of

44Am^{2}

. If the strip is bent in a semicircular shape, its magnetic moment will be_______

Am^{2}

(Given

\pi=\frac{22}{7}

)

Bending keeps the pole strength, not the moment

A bent magnet has the same pole strength but its poles are closer together, so its moment falls. A semicircle gives 2M/π, not M.

The equatorial field points opposite to the moment

On the axis the field of a short magnet points along M; on the equator it points opposite to M. The direction matters when this field is added to another.

Using the axial formula at an equatorial point

The factor 2 belongs to the axis only. Check where the point lies before choosing the formula: a point on the perpendicular bisector is equatorial.

Drill 5 more on magnetic dipole moment and the field of a bar magnet

Concept 2 of 2: Torque and potential energy of a dipole in a uniform field

A uniform field pulls the two poles of a magnet equally and oppositely, so there is no net force, only a torque that turns M towards B. The energy is lowest when M lies along B and highest when it points against B. The work needed to turn the dipole is the rise in this energy, so every work question is two values of −MB cos θ subtracted.

Definition

Torque: $\vec{\tau} = \vec{M} \times \vec{B}$ , size $MB\sin\theta$ , where θ is the angle between M and B.
Potential energy: $U = -\vec{M} \cdot \vec{B} = -MB\cos\theta$ . Stable at θ = 0 (U = −MB), unstable at θ = 180° (U = +MB).
Work to turn from $\theta_1$ to $\theta_2$ : $W = MB(\cos\theta_1 - \cos\theta_2)$ .
Stable to unstable: 2MB. Stable to 90°: MB. Stable to 60°: MB/2.
A torque at a known angle fixes the product MB: $MB = \tau/\sin\theta$ . Use it to find U or W without M or B separately.
A coil whose plane is perpendicular to B has its moment along B (θ = 0). A coil whose plane is parallel to B has θ = 90° and feels the largest torque.
Two dipoles at right angles to each other in one field: if one makes θ with B, the other makes 90° + θ, so equal torques mean $p_1\sin\theta = p_2\cos\theta$ .

Dipole in a uniform field

\tau = MB\sin\theta \qquad U = -MB\cos\theta \qquad W = MB(\cos\theta_1 - \cos\theta_2)

Worked example

A magnet of moment

3\ A\,m^{2}

lies at 30° to a uniform field of 0.2 T. Find the torque on it, its potential energy, and the work needed to turn it from the stable position to the most unstable one.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q76Moderate

Example 2 · Magnetism and Matter · Bar Magnets and Magnetic Dipoles

A magnetic dipole experiences a torque of

80\sqrt{3}\text{ }N\text{ }m

when placed in uniform magnetic field in such a way that dipole moment makes angle of

60^{\circ}

with magnetic field. The potential energy of the dipole is :

Dropping the minus sign in the energy

The potential energy is −MB cos θ. At angles below 90° it is negative, and an option with the right size but a positive sign is a deliberate trap.

Reading the plane of a coil as its moment

A coil's moment is along the normal to its plane. A plane perpendicular to B means the moment is along B, which is the stable position, not the position of largest torque.

Work is a difference of cosines, not of angles

Turning from 0° to 60° needs MB(1 − ½) = MB/2, not a third of the work for 0° to 180°. Always subtract the two values of cos θ.

Drill 7 more on torque and potential energy of a dipole in a uniform field

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Magnetic dipole moment and the field of a bar magnet
Short magnet: axial field, equatorial field, axial potential
$B_{axial} = \frac{\mu_0}{4\pi}\frac{2M}{r^{3}} \qquad B_{eq} = \frac{\mu_0}{4\pi}\frac{M}{r^{3}} \qquad V_{axis} = \frac{\mu_0}{4\pi}\frac{M}{r^{2}}$
Torque and potential energy of a dipole in a uniform field
Dipole in a uniform field
$\tau = MB\sin\theta \qquad U = -MB\cos\theta \qquad W = MB(\cos\theta_1 - \cos\theta_2)$

Watch out for (6)

Bending keeps the pole strength, not the moment
→ Magnetic dipole moment and the field of a bar magnet
The equatorial field points opposite to the moment
→ Magnetic dipole moment and the field of a bar magnet
Using the axial formula at an equatorial point
→ Magnetic dipole moment and the field of a bar magnet
Dropping the minus sign in the energy
→ Torque and potential energy of a dipole in a uniform field
Reading the plane of a coil as its moment
→ Torque and potential energy of a dipole in a uniform field
Work is a difference of cosines, not of angles
→ Torque and potential energy of a dipole in a uniform field

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