NEET 2027 · Physics · Class XII Chapter 4
Eleven pieces of school maths do all the heavy lifting in this chapter. Every one is explained in plain words, shown moving, and then used five times on real questions from the chapter itself — so you always know where it turns up.
Skill 01
Because the numbers here are either enormous or microscopic
An electron weighs 0.00000000000000000000000000000091 kg. Nobody writes that out. Instead we write 9.1 × 10−31 — a tidy front number, then a count of how many places the decimal point moved.
Once numbers are in that shape, the arithmetic becomes almost silly-easy, because you handle the front numbers and the little raised numbers separately.
Multiply three powers: 10⁻⁷ × 10 × 10⁻²
This is the top line of the Biot–Savart calculation in Example 4.4.
Biot–Savart · Example 4.4Work out qB for q = 1.6 × 10⁻¹⁹ C and B = 0.4 T
qB sits at the bottom of the radius, the period and the pitch formulas. Compute it once, reuse it three times.
Circular motionMultiply mv for m = 9 × 10⁻³¹ kg and v = 3 × 10⁷ m/s
Never leave 27 out front. Slide the point one place left and add one to the power. This is Example 4.3.
Radius r = mv/qB · Example 4.3Divide 2.7 × 10⁻²³ by 9.6 × 10⁻²³
When the powers are identical they vanish completely, and what is left is just a simple division.
Example 4.3 · radius of an electron's pathSquare a speed: (3 × 10⁷)²
Squaring DOUBLES the little number — it does not square it. Needed for every kinetic-energy line.
Kinetic energy ½mv²Adding is different from multiplying. 3 × 107 + 4 × 106 is not 7 × 1013. To add, first make both powers match: 3 × 107 + 0.4 × 107 = 3.4 × 107.
Skill 02
The quiet step that eats more marks than anything else
Every physics formula is built for SI units — metres, kilograms, amperes, tesla. But questions are written in whatever unit sounds natural: grams, centimetres, gauss, milliamps.
So before a single number goes into a formula, convert. Do it as a separate written step, not in your head.
A wire has mass 200 g. What goes into the formula?
Example 4.1, the floating wire. Leave it as 200 and your answer is a thousand times too big.
Example 4.1 · F = BILAn arc has radius 2.0 cm. Convert it.
Example 4.5, the semicircular arc carrying 12 A.
Example 4.5 · arc at the centreEarth-scale field of 6.5 gauss — in tesla?
Gauss is a smaller unit that survives in older books and lab equipment. Ten thousand gauss make one tesla.
Exercise 4.11 · electron in a chamberA solenoid 0.5 m long has 500 turns. Find n.
Example 4.8. Feeding 500 straight into B = μ₀nI is the classic solenoid mistake.
Example 4.8 · solenoidConvert 4.05 × 10⁻¹⁶ J into keV
Example 4.3. Joules are a clumsy unit for one particle — that is exactly why the electron-volt was invented.
Example 4.3 · energy of an electronConverting an area or a volume needs the conversion applied twice or three times. 1 cm² is not 0.01 m² — it is 10−4 m².
Skill 03
The dial that decides how much of a push actually counts
Almost every force in this chapter carries a sin θ on the end. That factor is a dial running from 0 to 1, and it decides what fraction of the maximum push you actually get.
The other job of trigonometry here is splitting a single arrow into two: the part going across something (that one uses sine) and the part going along it (that one uses cosine).
The values you must know cold
Notice sine climbs 0 → 1 while cosine falls 1 → 0. They are the same numbers read backwards.
Every force formulaA 0.5 m wire carries 2 A at 30° to a 0.4 T field. Find the force.
Without the sine you would have said 0.4 N — exactly double the truth.
F = BIL sin θA charge enters at 30° with speed 2 × 10⁶ m/s. Split it.
v⊥ sets the RADIUS of the helix. v∥ sets the PITCH. Swap them and both answers go wrong.
Helical motion · radius and pitchA coil starts with its normal along the field, then turns 90°.
Example 4.10. Here θ is measured from the NORMAL to the coil, not from its plane — the commonest slip in this whole chapter.
Example 4.10 · torque on a coilEarth's field runs south→north. A 1 A wire lies on a table.
Example 4.9. Line the wire up with the field and the magnet simply ignores it.
Example 4.9 · wire in Earth's fieldFor torque, θ is the angle between B and the normal to the coil. If a question says ‘the field lies in the plane of the coil’, that means θ = 90° and the torque is at its maximum, not zero.
Skill 04
How magnetism decides which way to push
Magnetism never pushes along a line you can guess. It pushes sideways to everything — and the cross product is the tool that says exactly which sideways.
It has two halves. The size is AB sin θ. The direction is at right angles to both input arrows, and you find it with your right hand.
A positive charge moves east; the field points north.
Reverse the charge, or swap the order, and it points down instead.
Lorentz forceThe unit-vector ring: î × ĵ = ?
Example 4.4 uses exactly this: an element along x, a point on the y-axis, so the field comes out along +z.
Example 4.4 · Biot–SavartA wire in a field: which way does it lurch?
Putting an arrow on I is a marked error even when the number is right.
F = Il × BWhy straight leads contribute nothing to an arc
Example 4.5(a). The straight bits are dead weight — only the curve makes any field at the centre.
Example 4.5 · semicircular arcTorque as a cross product
That aligned position is why a compass needle settles pointing north.
Torque on a current loopA × B = −B × A. Order matters, and swapping it flips the answer completely. Most sign errors in this chapter are really just this one mistake wearing a disguise.
Skill 05
Because √(something² + something²) is hiding in half the formulas
Whenever you stand off to one side of a wire or a loop, the distance to the far end is the slanted side of a right triangle. Pythagoras gives it to you: square the two legs, add, take the square root.
Examiners are kind here. They almost always pick numbers from the 3–4–5 family, so the square root comes out whole.
A wire sticks out 4 cm; you stand 3 cm from it. Find sin θ.
The discount factor for a finite wire drops straight out of the triangle. No calculator needed.
Finite straight wireA loop of radius 3 cm; a point 4 cm along the axis.
Once you see 3 and 4 in a loop question, the answer 5 is already waiting.
Field on the axis of a loopA wire 12 cm long; a point 8 cm away, level with its middle.
6-8-10 is just 3-4-5 doubled. Learn to spot the family, not the exact numbers.
Finite straight wireA point one radius out along a loop's axis (x = R).
Then cube it for the field: (0.707)³ = 0.354. The field has already fallen to about a third.
Loop axis at x = RTriangle families worth memorising
If a question gives you two legs and the third comes out ugly, check your reading — it usually means a slip somewhere.
All geometry questionsSkill 06
Loops, arcs and Ampèrian paths are all circles in disguise
Three facts about circles carry an astonishing amount of this chapter: the way round is 2πR, the area inside is πr², and a slice of a circle gives you exactly that fraction of the whole answer.
That last one is why arcs are so easy. A semicircle is half a loop, so it gives half the field. No new physics, just a fraction.
Adding every crumb of a circular loop
The 2π kills the 4π down to a 2, and one R cancels. That is the whole derivation.
Field at the centre of a loopAmpère's law on a straight wire
Two lines. Biot–Savart would need a page of integration for the same result.
Ampère's circuital lawA semicircular arc of radius 2.0 cm carrying 12 A
Example 4.5. Work out the whole circle first, then take your slice.
Example 4.5 · arcMagnetic moment of a 100-turn coil, radius 10 cm, 3.2 A
Example 4.10(b). The πr² is the only geometry in the whole magnetic moment.
Example 4.10 · magnetic momentHow much current sits inside radius r of a thick wire?
Example 4.7. This is why the field grows in a straight line as you move out from the axis.
Example 4.7 · inside a thick wireSkill 07
Including that alarming ³⁄₂ power on the loop-axis formula
A power of 3⁄2 looks frightening but is really two easy steps stuck together: take the square root, then cube it. Nothing more.
The other thing to feel in your bones is how brutal cubing is on a fraction. Cut something to 60% and cube it, and you are down to 21.6%.
Evaluate (x² + R²)^{3/2} for R = 3 cm, x = 4 cm
Root first, then cube. Never try to do the 3/2 in one go on a calculator under exam pressure.
Field on a loop's axisCube the ratio 0.6
Step one radius off the centre of a loop and this is what is left of the field — under a quarter.
Loop axis · shortcut methodSolve ½ℐω² = 20 J with ℐ = 0.1 kg·m²
Example 4.10(d). Whenever a squared unknown appears, isolate it first, then root at the very last step.
Example 4.10 · coil angular speedSquare 5.7 × 10⁷ m/s for a cyclotron energy
Squaring doubles the power of ten. It does not square it.
Cyclotron maximum energyThe r² underneath Biot–Savart
Example 4.4. Dividing by 0.25 is the same as multiplying by 4 — a useful sanity check.
Example 4.4 · Biot–SavartWatch which quantity carries the power. In the loop-axis formula the 3⁄2 applies to the whole bracket (x² + R²), never to x and R separately.
Skill 08
Answering ‘what happens if I double it?’ without doing any arithmetic
A huge number of questions never ask for a value at all. They ask what happens to one quantity when another changes. For those you do not need a calculator — you need to know which power of r you are dealing with.
This chapter contains three different fall-off rates, and telling them apart is worth real marks.
Move twice as far from a long straight wire.
The single most common error in this chapter is assuming an inverse square here. A whole wire is gentler than one element.
Long straight wireMove twice as far from one tiny current element.
Individually each crumb IS inverse-square. It is only the assembled infinite wire that comes out as 1/r.
Biot–Savart lawMove twice as far from a current loop, far away.
This ferocious fall-off is the signature of a dipole, and it is why an MRI is a tunnel rather than a magnet you stand beside.
Loop as a magnetic dipoleElectron and proton, same kinetic energy. Compare radii.
Equal SPEED would give r ∝ m and a ratio of 1/1836. Read which one the question said — it changes everything.
Circular motion · comparisonA meter reads 0.048 A when the true current is 1.00 A.
Example 4.12. That is why a bare galvanometer needs a shunt before it can be trusted as an ammeter.
Example 4.12 · galvanometerSkill 09
Where several of the chapter’s best results come from
Some of the most famous formulas here are not new physics at all. They are two things you already knew, set equal to each other, with a common factor crossed off.
The skill is spotting the factor that appears on both sides and striking it out — because what survives is usually the result the chapter is named for.
Where r = mv/qB comes from
No new law was used. Just circular motion plus the Lorentz force, joined at the hip.
Circular motionA wire floats in mid-air. Find B.
Example 4.1. Notice that only mass-per-metre really matters — the l partly cancels against the weight.
Example 4.1 · floating wireWhere B = μ₀nI comes from
The h vanishing is exactly why the answer does not depend on where inside the solenoid you stand.
SolenoidWhere v = E/B comes from
Because q cancelled and m never appeared, a velocity selector works on any particle at all.
Velocity selectorWhere T = 2πm/qB comes from
This cancellation is the entire reason a cyclotron can work at one fixed frequency.
Period · cyclotronSkill 10
Everything you need for the galvanometer, ammeter and voltmeter
Only the last part of the chapter needs circuit maths, but it needs it properly. Two rules cover it all.
In series, resistances simply add. In parallel, use R₁R₂/(R₁+R₂) — and the answer is always smaller than either one on its own. That smallness is the whole point of a shunt.
A 60 Ω galvanometer in a 3 V, 3 Ω circuit.
Example 4.12(a). The true current should be 1.00 A — the meter has strangled the circuit.
Example 4.12(a)Now shunt that galvanometer with 0.02 Ω.
Two resistors in parallel always give less than the smaller one. Here the 60 Ω has been rendered almost irrelevant.
Example 4.12(b)What does the shunted meter now read?
From 95% wrong to under 1% wrong, using one short fat piece of wire.
Example 4.12(c)Design a shunt: G = 100 Ω, full scale 1 mA, want 0–5 A.
Only 0.02% of the current goes through the delicate meter. The other 99.98% takes the bypass.
Ammeter conversionDesign a voltmeter: same meter, want 0–10 V.
An ammeter must end up with almost zero resistance; a voltmeter with almost infinite. Exact opposites.
Voltmeter conversionAn ammeter goes in series and must have low resistance. A voltmeter goes in parallel and must have high resistance. Mixing these up will short out a circuit or read nothing at all.
Skill 11
What that ∫ sign is actually asking you to do
The integral sign frightens people, but in this chapter it means something very ordinary: chop the thing into crumbs, work out what one crumb does, and add up all the crumbs.
You will almost never have to do a hard integral for NEET. What you need is to recognise what is being added, because in every case the chapter chooses shapes where the adding is trivial.
Adding round a circular loop
Three things that normally vary are all constant here, which turns a nasty sum into one multiplication.
Field at a loop's centreAdding round the solenoid's rectangle
Choosing a clever path is what makes Ampère's law usable. Three of the four sides were engineered to vanish.
SolenoidAdding round an Ampèrian circle
The symmetry is doing the work. Without it the sum stays a sum and tells you nothing.
Ampère's lawWork done as a coil swings through 90°
Example 4.10(d). Then ½ℐω² = 20 gives ω = 20 rad/s.
Example 4.10 · rotating coilA bent wire in a field
Same idea every time: small pieces, patiently added. That is all integration is here.
Force on a shaped conductor