Physics · Class 12 · Chapter 4
Everything a NEET aspirant needs on one page: force on a moving charge, Biot–Savart, Ampère's law, torque and the galvanometer — with live graphs to test the ideas.
Chapter 4 of Class 12 is the single largest source of direction-reasoning errors in the whole Physics paper. Roughly 2–3 questions appear every year, and they cluster in four places: force on a moving charge, Biot–Savart / Ampère field expressions, force between parallel wires, and galvanometer conversion.
Priority 1 — build first
Priority 2 — finish after P1
Your recurring pattern
Across ILTS papers the losses here were not formula recall — they were direction (equatorial field direction, drift direction) and ratio reversal (writing B1/B2 when the question asked B2/B1). Before you write the final option, re-read the asked line and check the order of the ratio.
T, ν and ω are independent of speed and radius. This is the single most-tested idea in the section: double the speed and the radius doubles, but the particle still takes exactly the same time per revolution.
Resolve: v⊥ = v sinθ makes the circle; v∥ = v cosθ is untouched and slides the circle along B.
NEET hook
A proton and an α-particle accelerated through the same potential V enter the same B. r = √(2mV/q)/B, so r ∝ √(m/q). For α: m × 4, q × 2 ⇒ rα/rp = √2. For the same momentum instead, r ∝ 1/q ⇒ ratio is 2:1. Read which quantity is held equal.
E, B and v mutually perpendicular. The electric force qE and magnetic force qvB oppose each other; only particles with one particular speed pass undeviated.
Syllabus gap
The cyclotron was removed from the rationalised NCERT text but is still listed in the NTA NEET syllabus and still appears in papers. Learn it from this box — you will not find it in your printed NCERT.
Two hollow D-shaped metal chambers (dees) sit in a strong perpendicular magnetic field, with a high-frequency alternating voltage across the gap. Inside a dee there is no electric field (electrostatic shielding), so the particle just travels a semicircle; every time it crosses the gap the polarity has reversed and it is accelerated again. Radius grows, but the time per semicircle does not.
Limitations — asked as one-liners
Definition of the ampere (old SI definition — still asked)
One ampere is the current which, in two infinitely long straight parallel wires of negligible cross-section placed 1 m apart in vacuum, produces a force of 2 × 10−7 N per metre of length on each wire.
dB is zero along the wire itself (θ = 0) and maximum perpendicular to it. Biot–Savart is to magnetism what Coulomb's law is to electrostatics — but with a cross product, so the field circles the current instead of pointing away from it.
| Configuration | Field | Where / notes |
|---|---|---|
| Infinite straight wire | B = μ0I / 2πr | B ∝ 1/r; circular field lines, right-hand grip rule |
| Finite straight wire | B = (μ0I/4πr)(sinφ1 + sinφ2) | φ measured from the perpendicular foot |
| Semi-infinite wire (at its end) | B = μ0I / 4πr | exactly half the infinite-wire value |
| Centre of circular loop | B = μ0NI / 2R | N turns; ⊥ to the plane of the loop |
| Arc of angle θ (radians) | B = μ0Iθ / 4πR | semicircle → μ0I/4R; quarter → μ0I/8R |
| Axis of a loop, distance x | B = μ0NIR² / 2(R²+x²)3/2 | x ≫ R ⇒ B ≈ μ02m/4πx³ (dipole) |
| Long solenoid (inside) | B = μ0nI | n = turns per metre; at either end, B = μ0nI/2 |
| Toroid (inside core) | B = μ0NI / 2πr | B = 0 in the hollow interior and outside |
Where it went wrong — direction
Two rules, do not mix them. Grip rule (thumb = current, fingers = field) gives the field produced by a current. F = qv × B / left-hand-free vector product gives the force felt by a charge. Papers punish you for using the grip rule where a cross product was needed. Also: for a negative charge the force reverses — write v × B first, then flip.
Valid for any closed (Amperian) loop; useful only where symmetry lets you pull B out of the integral. Currents outside the loop contribute to B at points on the loop but contribute nothing to the line integral.
Graph question you must recognise instantly
Solid wire: straight line rising from the origin up to r = a, then a 1/r decay. Hollow pipe: B = 0 everywhere inside (no enclosed current), a jump at r = a, then the same 1/r decay. Papers show the two graphs side by side as options.
A coil on a soft-iron core between concave pole pieces, so the field is radial — the plane of the coil is always parallel to B, which makes sinθ = 1 always and the scale linear.
Classic assertion–reason trap
Increasing the number of turns N increases current sensitivity — but the resistance Rg also rises in proportion, so voltage sensitivity need not increase. Statement true, reason true, but the reason does not explain the assertion. Assertion–reason items were lost on every ILTS paper; read both halves separately before deciding.
One-liners
An ideal ammeter has zero resistance and goes in series; an ideal voltmeter has infinite resistance and goes in parallel. A galvanometer can never be connected directly across a supply — the coil burns out.
| Quantity | Expression | Watch for |
|---|---|---|
| Magnetic force | F = qvB sinθ | zero work, speed constant |
| Radius / period | r = mv/qB; T = 2πm/qB | T independent of v and r |
| Pitch | p = 2πmv cosθ/qB | use cosθ, not sinθ |
| Velocity selector | v = E/B | independent of q and m |
| Cyclotron energy | K = q²B²R²/2m | gap content |
| Wire in a field | F = BIL sinθ | closed loop → net F = 0 |
| Parallel wires | F/L = μ0I1I2/2πd | same direction → attract |
| Straight wire field | B = μ0I/2πr | 1/r, not 1/r² |
| Loop centre / arc | μ0NI/2R; μ0Iθ/4πR | θ in radians |
| Axial field | μ0NIR²/2(R²+x²)3/2 | x = 0 recovers the centre value |
| Solenoid / toroid | μ0nI; μ0NI/2πr | half at a solenoid end |
| Torque / moment | τ = NIAB sinθ; m = NIA | θ from the normal |
| Galvanometer | SI = NAB/k; SV = NAB/kRg | the assertion–reason trap |
| Shunt / multiplier | S = IgRg/(I−Ig); R = V/Ig − Rg | convert mA to A first |
Final checklist before you hand in