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NCERT Class 12 · Physics · Chapter 5 · deep dive

The Bar Magnet,
Properly Understood

Four things, taken slowly and all the way down: where the axial and equatorial field formulas actually come from, why a magnet twists in a field, why it stores energy when you twist it, and what really changes when you cut one in half. Every idea is told twice — once as a story a ten-year-old can follow, once as the full derivation you need for the exam.

Drag the probeStep through the algebraSpin the magnetWatch the energy ballCut it two ways
PART 01

What a magnetic dipole actually is

Two ends, one number · the meaning of m
Story

The tug-of-war rope

Imagine a rope with a strong kid pulling at one end and an equally strong kid pulling at the other, in opposite directions. Two things decide how much twisting power this rope has: how strong the kids are, and how far apart they stand. Two strong kids standing close together do less twisting than two weaker kids standing far apart.

A bar magnet is exactly that rope. The "strength of each kid" is called the pole strength, written qm. The "distance apart" is the length of the magnet, 2l. Multiply them together and you get the one number that describes the whole magnet: its magnetic moment m.

m = qm × 2l    units: A m2 = J T−1
Maths

Why one number is enough

Everything a bar magnet does at a distance depends only on m, never on qm and l separately. A short fat magnet and a long thin one with the same product behave identically to anything far away. That is why NCERT writes every result in terms of m alone.

m is a vector. Its direction is defined as pointing from the south pole to the north pole inside the magnet — the same direction the field lines travel on their return journey through the material.

The same quantity describes a current loop: m = NIA. This is not a coincidence. Ampere's hypothesis says a bar magnet is a stack of tiny current loops, so the two definitions describe the same physical thing.

The two questions this whole topic answers

Question A: If I stand somewhere near a magnet, how strong is its field where I am standing? → that gives the axial and equatorial formulas, Parts 2 and 3.

Question B: If I put a magnet inside somebody else's field, what happens to it? → that gives torque and energy, Parts 4 and 5.

Keep these apart in your head. In Question A the magnet is the source. In Question B the magnet is the victim. Mixing them up is the commonest confusion in this chapter.

PART 02

The field along the axis

Standing in front of the magnet · deriving Baxial = (μ0/4π)(2m/r3)
Story

Two speakers, one facing you

Stand in a straight line out in front of the magnet's north end. Now there are two "speakers" playing at you: the north pole, which is nearer, and the south pole, which is further away.

The north pole pushes the field away from itself, so it pushes outward, towards you. The south pole pulls the field towards itself, so it pulls inward, away from you. They are fighting.

But the near speaker is louder. The north pole wins, and what is left over is the field you feel. Because the loser is not far behind, most of the effect cancels — which is why the leftover fades away very fast as you walk backwards.

Lab 1 · Build the axial formula, one line at a time

tap next

The diagram highlights whichever piece the current step is talking about. Do not skip — each line follows from the one above it.

The step everyone rushes

The algebra hinges on one identity: (r+l)² − (r−l)² = 4rl. Expand both squares and almost everything cancels — the r² terms and the l² terms both disappear, leaving only the cross terms. This is why the final answer has a single power of r on top before the approximation is made.

Where "short magnet" comes in, and why it matters

The exact answer is B = (μ0/4π) × 2mr/(r² − l²)². That is true for any distance. Only when r is much bigger than l do we say r² − l² ≈ r² and get the familiar 2m/r³.

So "short bar magnet" in an exam question is not decoration. It is permission to drop the l². Use Lab 2 below to watch how big the error is when that permission is stretched.

Lab 2 · Exact versus approximate

drag the distance

Both formulas are computed live for the same magnet. Watch the gap between them close as you move the probe away.

r = cm l = cm Exact: Short-magnet: Error:

Rule of thumb: once r is about five times l, the short-magnet formula is within roughly 2%. Below that it is noticeably wrong — which is exactly why NEET always says "short bar magnet".

PART 03

The field on the equator

Standing beside the magnet · deriving Beq = −(μ0/4π)(m/r3)
Story

Two ropes pulling sideways

Now walk round and stand beside the magnet, level with its middle. Something different happens. Now both poles are exactly the same distance from you, so neither one is louder.

Each pole tugs the field along a slanting line towards or away from itself. Break each tug into two parts: a part pointing sideways (towards or away from the magnet) and a part pointing along the magnet.

The sideways parts point in opposite directions and cancel perfectly. The along-the-magnet parts point the same way and add up. So the leftover field lies flat along the magnet's direction — and it points backwards, from N towards S. That backwards-ness is the minus sign in the formula.

Lab 3 · Build the equatorial formula

watch the components

The diagram shows the two pole contributions and splits them into components as the derivation proceeds.

Lab 4 · Both places at once

drag the probe

Drag the orange probe anywhere. The purple probe automatically mirrors it onto the equatorial line at the same distance, so you are always comparing like with like.

Distance r = cm Baxial = Bequatorial = Ratio =

The arrows show direction, not just size. Notice that the axial arrow always points the same way as m, and the equatorial arrow always points the opposite way — whatever distance you choose.

Axis is twice the equator — and here is exactly why

On the axis, the near pole beats the far pole and the leftover is what survives a subtraction. On the equator, the two poles are equal and half of each contribution is thrown away by cancellation.

Run both derivations to the end and the numerator is 2m in the first case and m in the second. Same r3, same constant. So the ratio is exactly 2 : 1, at every distance, for every magnet.

Quick check

A short magnet's moment m points along +x. At a point on the equatorial line, the field it produces points along

PART 04

Torque — why a magnet twists

τ = m × B · and why the net force is exactly zero
Story

Two hands on a steering wheel

Put your left hand on the left of a steering wheel and push up. Put your right hand on the right and push down. Equal pushes, opposite directions.

Does the wheel move across the car? No — the two pushes cancel, so there is no net force. Does the wheel turn? Yes — because the two pushes are not on the same line, they twist it.

A magnet in a uniform field is exactly this. The field pushes the north pole one way and the south pole the opposite way, with equal strength. No net force, but a definite twist. Physicists call a pair of forces like this a couple.

Maths

Deriving τ = mB sin θ in four lines

Line 1. Force on the north pole is qmB along B. Force on the south pole is qmB opposite to B. Same size, opposite direction ⇒ net force = 0.

Line 2. Torque of a couple = (one force) × (perpendicular distance between the two lines of action).

Line 3. If the magnet makes angle θ with B, that perpendicular distance is 2l sin θ. (When θ = 0 the two forces sit on the same line and the distance is zero.)

Line 4. So τ = qmB × 2l sin θ = (qm × 2l) B sin θ = mB sin θ.

τ = m × B    |τ| = mB sin θ    Fnet = 0

The distinction worth four marks

Zero net force is a consequence of the field being uniform. Both poles sit in the same field strength, so their forces match exactly and cancel.

In a non-uniform field the nearer pole sits where the field is stronger, so the forces no longer match and a net force F = m(dB/dx) survives. That leftover is the entire reason a magnet can pick up a paperclip. Uniform ⇒ turn only. Non-uniform ⇒ turn and tug.

PART 05

Energy — the hill and the valley

U = −m·B · where the minus sign comes from
Story

A ball in a bowl

Rolling a ball up the side of a bowl takes effort, and the ball stores that effort. Let go and it rolls back down. The bottom of the bowl is the comfortable place: lowest energy, and the ball happily stays there.

A magnet in a field lives in a bowl too, except the bowl is made of angles instead of distance. Lined up with the field is the bottom of the bowl. Turning it away is rolling uphill, and you have to supply the energy.

Turn it all the way to backwards and something odd happens: it balances. That is the very top of the hill. It will sit there if you are careful — but the tiniest nudge and it tumbles all the way back down.

Maths

Integrating the torque

To turn the magnet by a small angle dθ against the restoring torque, the work you must do is dW = τ dθ = mB sin θ dθ.

Add up all the small bits: U = ∫ mB sin θ dθ = −mB cos θ + C.

NCERT takes the constant of integration C to be zero. That choice places the zero of energy at θ = 90°, when the magnet sits square across the field. So finally U = −mB cos θ = −m·B.

The minus sign is not decoration — it is what makes the aligned position a valley (U = −mB, minimum) rather than a peak.

Lab 5 · The linked twist-and-energy simulator

drag, then release

Drag the magnet on the left to any angle. The ball on the right sits at the matching point of the energy curve — they are the same object seen two ways. Then press Release and watch what the physics does on its own.

θ = τ = mB sinθ = U = −mB cosθ = State:

Using m = 1 and B = 1 so the numbers are simply sinθ and −cosθ. Try the hilltop button: leave it exactly at 180° and it stays; nudge it a degree and it falls. That is what "unstable equilibrium" means.

Three landmarks worth memorising

Angle θTorque τEnergy UWhat it is
0−mB (minimum)Stable equilibrium — bottom of the valley
90°mB (maximum)0Hardest twist, zero stored energy
180°0+mB (maximum)Unstable equilibrium — top of the hill

Read across the middle row carefully. Torque and energy never peak at the same angle. Sketch sinθ and −cosθ on one axis once and this never confuses you again.

Work done between any two angles

W = U2 − U1 = mB(cos θ1 − cos θ2)

Starting from aligned (θ1 = 0), three cases cover almost every exam question: to 60° costs mB/2, to 90° costs mB, to 180° costs 2mB. Compute mB once and read all three off.

Quick check

At which angle is the torque on a magnet maximum and its potential energy exactly zero?

PART 06

Cutting a magnet — two ways

What survives the scissors, and what does not
Story

Chocolate bar, two ways to break it

Think of the magnet as a chocolate bar. You can snap it across the middle, getting two short bars. Or you can slice it lengthways, getting two long thin bars.

Snapping across makes each piece shorter, but each piece is still just as thick. Slicing lengthways keeps the full length, but each piece is now thinner.

Here is the thing to hold onto: pole strength depends on how thick the end face is. A fat end is a strong pole; a thin end is a weak pole. So the two cuts damage different things — and yet, remarkably, they damage the magnetic moment by exactly the same amount.

Lab 6 · The cutting bench

pick a cut

Choose a cut and watch which quantities move. Grey means unchanged; red means halved.

QuantityBeforeAfter (each piece)Verdict

Pick a cut above to see what changes.

Maths

Both cuts, worked through

Transverse cut (across the length). The end faces are untouched, so pole strength qm stays the same. The length halves: 2l → l. Therefore

m′ = qm × l = (qm × 2l)/2 = m/2

Longitudinal cut (along the length). The length is untouched, so 2l stays the same. But the cross-sectional area halves, so each new end face is half as big and qm → qm/2. Therefore

m′ = (qm/2) × 2l = m/2

Same answer, opposite reasons. If a written question asks you to explain, the reason is what earns the mark, not the number.

The one thing that never changes, no matter how you cut

Every piece, however small, is a complete magnet with its own north and south pole. You never obtain a lone pole. Keep cutting down to a single atom and that atom is still a tiny current loop — still a dipole.

This is why Gauss's law for magnetism reads φB = 0 with nothing on the right-hand side. There is no isolated source to count. Free poles obtained by cutting: always zero.

Bonus — what happens to the swinging period

If the magnet is hung so it can oscillate, the period is T = 2π√(I/mB). The two cuts now behave differently, because they affect the moment of inertia differently.

CutMoment mInertia INew period T′
Transversem/2I/8T/2
Longitudinalm/2I/2T (unchanged)

Why I falls by eight in the transverse case: I = MbL2/12, and both the mass and the length halve, so I drops by 2 × 4 = 8. In the longitudinal case only the mass halves, so I drops by 2 — matching the drop in m exactly, and the period does not budge.

Note: this period formula is not in your rationalised NCERT text, though NEET still asks it. Treat it as gap content.

Quick check

A bar magnet is cut into 6 pieces. How many free (isolated) magnetic poles are produced?

PART 07

The whole topic on one page

Everything above, compressed
m = qm × 2l = N I A   (direction: S → N inside the magnet) Baxial = (μ0/4π)(2m/r3)   along m   [exact: 2mr/(r²−l²)²] Bequatorial = −(μ0/4π)(m/r3)   against m   [exact: m/(r²+l²)3/2] τ = m × B, |τ| = mB sinθ   Fnet = 0 in a uniform field U = −m·B = −mB cosθ   W = mB(cosθ1 − cosθ2) Either cut ⇒ m′ = m/2   free poles = 0   always

Five sentences that carry the whole topic

1. Axis-along, equator-against — and the axis is twice as strong.
2. A uniform field can only turn a magnet, never tug it.
3. Torque peaks at 90°; energy peaks at 180°; they never peak together.
4. Aligned is the valley (−mB), backwards is the hilltop (+mB), and the climb between them costs 2mB.
5. Both cuts halve the moment, for opposite reasons, and neither ever frees a pole.

If you only rehearse one thing

Rehearse Part 5. Torque and energy questions are, by a wide margin, the most-asked calculation in this chapter's entire past-paper history — and nearly all of them reduce to τ = mB sinθ or W = 2mB. If you can do those two without thinking, a large share of the chapter's marks is already secure.

Then rehearse the direction facts from Parts 3 and 4. Those are where marks leak away silently — you can get the number right and the sign wrong, and the answer key does not care which mistake you made.