Class 12 Physics · NCERT notes for NEET

Magnetism and Matter: NCERT Notes, Formulas and Tricks for NEET

The bar magnet as a magnetic dipole (field lines, axial and equatorial fields, torque, energy, Gauss's law) and how dia-, para- and ferromagnetic materials respond to a field. It is a short chapter; NEET usually asks one question from it, most often on the classification of materials or the dipole formulas.

1. Bar magnet as a magnetic dipole
2. Magnetic properties of materials 🔒

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Topic 1

Bar magnet as a magnetic dipole

Break a fridge magnet into ten pieces and you get ten complete magnets, each with its own N and S: nobody has ever found a lone magnetic pole.

A freely suspended magnet settles roughly along north-south. Like poles repel, unlike poles attract.
Poles always come in pairs: cutting a magnet gives two smaller magnets. Isolated magnetic monopoles are not known.
Field lines are continuous closed loops: outside the magnet they run from N to S, inside from S to N. They never cross; the tangent gives the direction of $B$ and crowding shows strength.
A bar magnet and a current-carrying solenoid have the same external field pattern, so a bar magnet can be treated as an equivalent solenoid. A current loop has moment $m = I A$ .
Magnetic moment $m$ points from S to N. SI unit $A m^{2}$ , which is the same as $J T^{- 1}$ .
For a short magnet ( $r ≫ l$ ): axial field $= \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}}$ along $m$ ; equatorial field $= \frac{μ _{0}}{4 π} \frac{m}{r ^{3}}$ opposite to $m$ . So $B_{a x ia l} = 2 B_{e q}$ at equal distance.
In a uniform field there is no net force on a magnet, only a torque $τ = m \times B$ that turns $m$ towards $B$ .
Potential energy $U = - m \cdot B$ : minimum ( $- m B$ , stable) at $θ = 0^{\circ}$ and maximum ( $+ m B$ , unstable) at $θ = 18 0^{\circ}$ .
A magnet slightly displaced in a uniform field oscillates in SHM with $T = 2 π I / (m B)$ ; this is how $m$ or $B$ is measured.
Gauss's law for magnetism: the net magnetic flux through any closed surface is zero, $\oint B \cdot d S = 0$ , because there are no monopoles.

Each line leaves N, curves round to S and returns through the magnet from S to N, closing the loop. Inside lines show the direction of m. — Each line leaves N, curves round to S and returns through the magnet from S to N, closing the loop. Inside lines show the direction of $m$ .

At the same distance r, the axial field is twice the equatorial field. Axial B is along m; equatorial B is opposite to m. — At the same distance $r$ , the axial field is twice the equatorial field. Axial $B$ is along $m$ ; equatorial $B$ is opposite to $m$ .

Energy is lowest at 0^circ and highest at 180^circ, where torque is zero. Torque peaks at 90^circ. Turning from 0^circ to 180^circ costs 2mB. — Energy is lowest at $0^{\circ}$ and highest at $18 0^{\circ}$ , where torque is zero. Torque peaks at $9 0^{\circ}$ . Turning from $0^{\circ}$ to $18 0^{\circ}$ costs $2 m B$ .

Formulas

$B_{a x ia l} = \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}}$
On the axis of a short magnet, along $m$
$B_{e q} = \frac{μ _{0}}{4 π} \frac{m}{r ^{3}}$
On the equatorial line, opposite to $m$
$B_{a x ia l} = \frac{μ _{0}}{4 π} \frac{2 m r}{( r ^{2} - l ^{2} ) ^{2}}$
Exact axial field, $2 l$ = magnetic length
$m = q_{m} (2 l)$ , $m = I A$
Pole strength times magnetic length; current loop
$τ = m \times B$ , $τ = m B sin θ$
Torque in a uniform field
$U = - m \cdot B = - m B cos θ$
Potential energy of the dipole
$W = m B (cos θ_{1} - cos θ_{2})$
Work to rotate from $θ_{1}$ to $θ_{2}$
$T = 2 π \frac{I}{m B}$
Small oscillations; $I$ = moment of inertia
$\oint B \cdot d S = 0$
Gauss's law for magnetism

Tips, tricks and hacks

Copy from electrostaticsEvery dipole result carries over with

p \to m

E \to B

and

\frac{1}{4 π ε _{0}} \to \frac{μ _{0}}{4 π}

. So

E_{a x ia l} = \frac{2 p}{4 π ε _{0} r ^{3}}

becomes

B_{a x ia l} = \frac{μ _{0}}{4 π} \frac{2 m}{r ^{3}}

;

τ = pE sin θ

becomes

m B sin θ

2-to-1 and cube ruleAxial : equatorial

= 2 : 1

at the same

r

, and both fall as

1/ r^{3}

. Example:

B_{e q} = 2 μ T

at 10 cm, so

B_{a x ia l} = 4 μ T

at 10 cm and

4/8 = 0.5 μ T

at 20 cm.

Work from 0 degreesStarting from equilibrium,

W = m B (1 - cos θ)

6 0^{\circ} \to m B /2

9 0^{\circ} \to m B

18 0^{\circ} \to 2 m B

. Example:

m = 2 A m^{2}

B = 0.5

T, turn to

9 0^{\circ}

W = 1

Cut magnet, new periodCut a bar magnet into two equal halves across its length: each half has

m /2

and

I /8

(half mass, half length). So

T^{'} = 2 π (I /8) / (m B /2) = T /2

Common mistakes

✗ Field lines go from N to S everywhere, including inside the magnet.

✓ Inside they go S to N; that is what makes them closed loops with zero net flux.

✗ Torque is maximum when the magnet is along the field.

✓ At $θ = 0^{\circ}$ torque is zero (equilibrium); it is maximum at $9 0^{\circ}$ .

✗ The equatorial field points along $m$ , like the axial field.

✓ On the equatorial line $B$ is antiparallel to $m$ , while on the axis it is parallel.

✗ A closed surface around only the N pole has non-zero flux.

✓ You cannot enclose only one pole: lines that leave also return through the magnet, so flux is zero.

NEET traps

"A magnet in a uniform field experiences a force and a torque": false, only a torque; a non-uniform field is needed for a net force.

Both $A m^{2}$ and $J T^{- 1}$ are correct units of $m$ ; NEET may offer one of them as the "wrong" option.

Work $0^{\circ} \to 18 0^{\circ}$ is $2 m B$ , but work $9 0^{\circ} \to 18 0^{\circ}$ is only $m B$ ; read the starting angle.

Quick check: A short bar magnet gives

4 \times 1 0^{- 6}

T at a point on its axis 10 cm from its centre. The field at a point on its equatorial line 20 cm from the centre is

$5 \times 1 0^{- 7}$ T
$2.5 \times 1 0^{- 7}$ T
$1 \times 1 0^{- 6}$ T
$2 \times 1 0^{- 6}$ T

Answer: B. Equatorial at 10 cm $= 4/2 = 2 \times 1 0^{- 6}$ T. Doubling $r$ divides by 8: $2.5 \times 1 0^{- 7}$ T.

Topic 2 · 🔒

Magnetic properties of materials

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