The Twist and the Valley · Dipole in a Uniform Field

PART 01

What "uniform field" actually promises

The one assumption everything else rests on

Story

Rain falling perfectly evenly

Imagine rain falling straight down, exactly the same everywhere — same speed, same direction, no gusts, no heavier patches. Hold out a stick and every part of that stick gets hit identically. Nothing about your position changes what the rain does to you.

That is what uniform field means. Same strength, same direction, at every point. And it has one enormous consequence: because the north end and the south end of your magnet sit in identical conditions, the pushes they receive are identical in size.

Identical sizes, opposite directions. That is the entire story of this topic. It gives you a twist and no push — and the reason will become obvious in Part 3.

The cast of characters

Symbol	Name	What it is	Unit
m	Magnetic moment	Strength × length of the dipole; points S → N inside it	A m² = J T⁻¹
q_m	Pole strength	How strong one end is on its own; m = q_m × 2l	A m
B	Magnetic field	The external field the dipole sits in	T
θ	Angle	Between m and B — never between anything else	degrees or rad

The θ trap. Every formula on this page measures θ from m to B. If a question gives you the angle between the magnet and something else — the vertical, the coil's plane, the normal — convert it first. More marks are lost to this than to any actual physics.

PART 02

Deriving the torque

τ = mB sin θ from two forces and a perpendicular distance

Story

Two hands on a steering wheel

Left hand on the left of the wheel, push up. Right hand on the right, push down. Equal pushes, opposite directions.

Does the wheel slide across the car? No — the pushes cancel. Does it turn? Yes — because they are not on the same line.

Now notice something. If you moved both hands to the top of the wheel and pushed one up and one down, they would be almost on the same line, and almost nothing would happen. How much turning you get depends on how far apart the two lines of push are — and that distance changes as the magnet rotates. That changing distance is where sinθ comes from.

Lab 1 · Build the couple

step through it, then drag the angle

Each step adds one layer to the picture. The angle slider works at every step, so you can watch each new piece respond as the magnet turns.

Angle θ

θ = — Force on each pole = q_mB = — Perpendicular distance = 2l sinθ = — τ = —

Drag the angle to 0° and watch the perpendicular distance collapse to nothing — the two forces line up and the twist vanishes. Drag to 90° and it reaches its maximum, 2l.

τ = m × B |τ| = mB sin θ

Where each piece of the formula came from

The m came from q_m × 2l — pole strength times length, collapsing into one symbol.
The B came from the force on each pole, q_mB.
The sinθ came from the geometry — the perpendicular distance between the two lines of action is 2l sinθ, not 2l.

That third one is the piece students forget. The forces do not change size as the magnet turns; only their separation does.

PART 03

Why the net force is exactly zero

Not approximately zero · exactly zero

Maths

The two-line proof

Model the dipole as poles +q_m and −q_m, separated by 2l.

Force on the north pole: +q_mB, along the field.
Force on the south pole: −q_mB, against the field.

Because the field is uniform, B has the same value at both ends. So the two forces have identical magnitude q_mB and opposite directions:

F_net = q_mB + (−q_mB) = 0 for every angle θ

They do not share a line of action, so they form a couple — pure rotation, no translation.

The distinction that is worth four marks

Zero net force is a consequence of uniformity, nothing else. In a non-uniform field the nearer pole sits where the field is stronger, so the forces no longer match and a net force survives:

F = m (dB/dx) (non-uniform field only)

That leftover is the entire reason a magnet can pick up a paperclip. A uniform field could only ever turn the clip, never lift it.

One line to carry: uniform ⇒ turn only. Non-uniform ⇒ turn and tug.

Quick check

A bar magnet sits at 37° in a uniform field. The net force on it is

PART 04

Which way does the torque point?

τ = m × B is a vector · order matters

Story

Tightening a screw

When you turn a screwdriver clockwise, the screw goes into the wall. Turn it anticlockwise and it comes out. The turning happens in a flat circle, but the result points along an axis perpendicular to that circle.

Torque works the same way. The magnet turns in a flat plane, but the torque vector points straight out of that plane — either towards you or away from you. Which one tells you which way it is turning.

Lab 2 · The cross product, seen properly

drag m, swap the order

Drag the red moment arrow. The symbol at the centre shows the torque vector: ⨀ means out of the page (towards you), ⨂ means into the page. Its size shows the magnitude.

Angle θ

Computing: m × B θ = — |τ| = — Direction: — Effect: —

Press the swap button and every direction flips, while every magnitude stays the same. That is what "order matters in a cross product" means — and why writing B × m gives an answer that is right in size and wrong in sign.

Maths

Doing it with unit vectors

Put B along +x and let m lie in the xy-plane at angle θ:

m = m(cosθ î + sinθ ĵ) B = B î

Then, using î × î = 0 and ĵ × î = −k̂:

τ = m × B = mB sinθ (ĵ × î) = −mB sinθ k̂

The minus sign is doing real physical work. If θ is positive (m rotated anticlockwise from B), the torque points into the page, which rotates m clockwise — back towards B. The torque is always restoring. That single sign is why a compass needle settles rather than spinning away.

Memory aid for the cycle: î → ĵ → k̂ → î. Going forwards round the cycle gives +, backwards gives −.

PART 05

Energy — and where the minus sign comes from

U = −m·B by integrating the torque

Story

Rolling a ball up the side of a bowl

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

A magnet lives in a bowl too, except the bowl is made of angles rather than distance. Lined up with the field is the bottom. Turning it away is rolling uphill.

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

Maths

The integration, with the constant explained

To turn the dipole through a small angle dθ against the restoring torque, you must do work dW = τ dθ = mB sinθ dθ. Adding up all the small bits:

U = ∫ mB sinθ dθ = −mB cosθ + C

Now the constant. NCERT sets C = 0, which places the zero of energy at θ = 90° — the dipole square across the field. Why choose that? Because it makes the formula collapse into the elegant dot product:

U = −m · B = −mB cos θ

The minus sign is not decoration. It is what makes the aligned position a valley (U = −mB, minimum) rather than a peak. Drop the minus and the physics inverts — the compass would flee the field instead of following it.

Note which product each formula uses

Torque uses a cross product: τ = m × B. The result is a vector, and it is largest when m and B are perpendicular.

Energy uses a dot product: U = −m · B. The result is a scalar, and it is largest in magnitude when m and B are parallel or antiparallel.

Cross and dot are 90° out of step with each other — which is exactly why torque and energy never peak at the same angle. The whole of Part 6 is that one sentence, made visible.

PART 06

The two curves, side by side

Why τ and U never peak together

Lab 3 · Torque and energy on one axis

drag or release

Drag the magnet on the left. Both curves on the right carry a marker at the current angle. Press Release and the physics takes over — watch the markers sweep along together.

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

Set θ = 90° and read both numbers: torque is at its maximum of 1, energy is exactly 0. Now set θ = 180°: torque has fallen to 0 while energy has climbed to its maximum of +1. The two are never large at the same moment.

θ	τ = mB sinθ	U = −mB cosθ	Name	What happens if nudged
0°	0	−mB (minimum)	Stable equilibrium	Restoring torque pushes it back
60°	0.87 mB	−0.5 mB	—	Turns back towards 0°
90°	mB (maximum)	0	Maximum twist	Turns back towards 0°, hardest push
120°	0.87 mB	+0.5 mB	—	Turns back towards 0°
180°	0	+mB (maximum)	Unstable equilibrium	Falls all the way to 0°

Quick check

At which angle is the torque maximum and the potential energy exactly zero?

PART 07

Work done in rotating

W = mB(cosθ₁ − cosθ₂) · the most-asked calculation in the chapter

Maths

It is just a height difference

Work done by an external agent equals the increase in potential energy. Nothing more:

W = U₂ − U₁ = (−mB cosθ₂) − (−mB cosθ₁) W = mB(cos θ₁ − cos θ₂)

Note the order carefully: the starting angle comes first. Written this way there is no stray minus sign to lose, which is why this form is safer under time pressure than W = −mB(cosθ₂ − cosθ₁).

Think of the U-curve as a hillside and the two angles as two points on it. The work is simply how much higher the second point is than the first. If it comes out negative, you went downhill and the field did the work for you.

Lab 4 · The energy ladder

pick two angles

Set a start and an end angle. The shaded height on the energy curve is the work done. Try the preset buttons for the three cases that appear again and again in exams.

Start θ₁

End θ₂

U₁ = — U₂ = — W = U₂ − U₁ = — Meaning: —

All values are in units of mB. Press the last preset to see a negative work — going from the hilltop down to the valley, the field does the work and you get energy back.

The three cases worth memorising outright

Rotation	Work	How to see it instantly
0° → 60°	mB/2	cos goes 1 → ½, a drop of ½
0° → 90°	mB	cos goes 1 → 0, a drop of 1
0° → 180°	2mB	cos goes 1 → −1, a drop of 2

Compute mB once, then read all three off. And learn the phrase-to-formula map, because exams vary the wording constantly: "stable to unstable", "parallel to antiparallel", "0° to 180°", "most stable to most unstable" all mean the same thing — 2mB.

The two-in-one shortcut

Many questions give you a torque and ask for work, without ever mentioning m or B separately. Do not find them. Extract the product instead:

mB = τ/sinθ ⇒ W_{stable→unstable} = 2mB = 2τ/sinθ

JEE Main 2020 gives τ = 0.018 N m at θ = 30°, so W = 2(0.018)/0.5 = 0.072 J in a single line. JEE Main 2026 asks the identical question with τ = 0.016 N m at 30° → W = 0.064 J. Treat mB as one unknown "package" and never split it.

PART 08

Why a compass always finds north

The restoring torque, and what it leads to

Story

A ball at the bottom of a bowl, again

Whatever angle you release the needle at, the torque always points back towards alignment. Never away. So the needle swings towards θ = 0°, overshoots (it has picked up speed), swings back, overshoots less, and eventually — because of a little friction — settles.

That settling is why a compass works at all. The needle is not being told where north is. It is simply rolling to the bottom of an energy bowl, and the bottom happens to face north.

Maths

Why the swinging is simple harmonic

For a small displacement θ, use sinθ ≈ θ. The restoring torque becomes

τ = −mB sinθ ≈ −(mB)θ

Torque proportional to displacement, opposite in sign — the exact signature of simple harmonic motion, with mB playing the role of the "spring constant".

Newton's second law for rotation, Iα = −(mB)θ, then gives ω² = mB/I and hence

T = 2π√(I / mB)

Syllabus note: this period formula was removed from the rationalised NCERT, though NEET still sets it — treat it as gap content. The qualitative statement that the needle oscillates and settles is fully in syllabus.

Reading the equilibria properly

Both θ = 0° and θ = 180° have zero torque, so both are equilibria. The energy decides which is which.

Two conditions, applied in order: zero torque picks the candidates (0° and 180°); minimum energy picks the winner (0°). Testing only one condition is how students end up calling 180° stable — it genuinely is an equilibrium, just not a stable one.

PART 09

Drill it until it is automatic

Fresh problems every time you press the button

Lab 5 · The problem generator

unlimited practice

A new problem every press, drawn from the four types that account for nearly every exam question on this topic. Type your answer and check it — the worked solution appears either way.

Attempted: 0 Correct: 0 Accuracy: — Target: under 40 seconds each

Answers are accepted within 2% to allow for rounding. If you are consistently slower than about forty seconds on these, that is the single most valuable thing to fix in this chapter.

PART 10

Everything on one page

The whole topic, compressed

τ = m × B · |τ| = mB sinθ · τ_max = mB at θ = 90° F_net = 0 in a uniform field · F = m(dB/dx) if non-uniform U = −m·B = −mB cosθ · U_min = −mB at 0° · U_max = +mB at 180° W = mB(cosθ₁ − cosθ₂) · 0→90°: mB · 0→180°: 2mB

Six sentences that carry the whole topic

1. θ is always measured between m and B — convert any other angle first.
2. Uniform field gives a couple: zero net force, non-zero torque, at every angle.
3. The sinθ comes from geometry — the perpendicular distance 2lsinθ, not from the forces changing.
4. Torque is a cross product (vector, peaks at 90°); energy is a dot product (scalar, peaks at 0° and 180°). They are 90° out of step.
5. Aligned is the valley (−mB), backwards is the hilltop (+mB), and the climb between them is 2mB.
6. The torque is always restoring, which is why a free dipole ends up aligned — and why a compass works.

Five traps, in the order they cost marks

1. Using the wrong angle — the plane's angle instead of the normal's, or the magnet's angle to the vertical.
2. Answering the intermediate value. If a question needs m = NIA then torque, the moment alone is always planted as an option.
3. Forgetting to double for the 0° → 180° case. mB is the 90° answer; 2mB is the 180° answer.
4. Dropping the minus sign in U. Stable equilibrium energy is negative; if yours came out positive you used the wrong angle.
5. Claiming a net force exists in a uniform field. It is exactly zero, always.

Why this page matters more than the rest of the chapter

Going through the full NEET/AIPMT record from 2001 and the JEE Main record from 2003, torque and energy questions account for roughly a fifth of everything ever set from Magnetism and Matter — and nearly all of them reduce to τ = mB sinθ or W = 2mB.

AIPMT 2009, AIPMT 2011, AIPMT 2012, AIEEE 2003, NEET 2016, NEET 2024, JEE Main 2020, 2022, 2023, 2024, 2025 and 2026 are all the same two formulas in different clothes. No other topic in this chapter repeats like that.

So the target is not "can I do it" but "can I do it in under forty seconds without writing much down". Use the generator in Part 9 until that is true, and a large share of this chapter's marks is secured before you revise anything else.