🏠 NEET Home
AITS 03Physics · Chapter 3 of 6

Physics · last-minute revision

Moving Charges and Magnetism

Force on charges and wires, Biot–Savart and Ampère results, torque on a coil, galvanometer conversions and the cyclotron — with every numerical in the check verified.

F = q(v × B)every result starts from the cross product

Force on a moving chargeMust

Lorentz force
F = q[E + v × B]; magnetic part F = qvB sinθ
Unit of B
tesla (T) = N s C−1 m−1; 1 gauss = 10−4 T
Zero magnetic force when
q = 0, v = 0, or v ∥ B (θ = 0° or 180°)
Direction
Fleming's left-hand rule (for +q; reverse for electron)
Charged particles in circular orbits in a magnetic field with equal periods×××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××++B into pagePositive charge, v ⊥ Bbrass = velocity, red = force (towards centre)r = mv/qB = p/qB = √(2mK)/qBT = 2πm/qBω = qB/m, f = qB/2πmBoth charges finish a lap together:T does not depend on v or r.Magnetic force ⊥ v → no work, speed constant.+ve charge in B into page goes anticlockwise.
Two positive charges in the same B complete a revolution in the same time. Faster charge → larger circle.
Helical path of a charge entering a magnetic field at an anglepitch pBr = mv sinθ / qBp = 2πm v cosθ / qBθ = angle between v and B. θ = 0° → straight line; θ = 90° → circle; otherwise → helix along B.
Parallel component v cosθ carries the charge along B; perpendicular component v sinθ makes the circle.

Radius ratios: decide what is held equal

Held equalr depends onp : d : α (p = proton, d = deuteron, α = alpha)
accelerating voltage Vr = √(2mV/q)/B ∝ √(m/q)1 : √2 : √2
kinetic energy Kr = √(2mK)/qB ∝ √m / q1 : √2 : 1
momentum pr = p/qB ∝ 1/q1 : 1 : 1/2
speed vr = mv/qB ∝ m/q1 : 2 : 2
Velocity selector (crossed E, B)
undeflected if qE = qvB → v = E/B, independent of q and m

Force on a current-carrying conductorMust

Straight wire
F = I L × B, F = BIL sinθ
Any shape in uniform B
F = I (Leff × B), Leff = straight line from start to end point
Closed loop in uniform B
net force = 0 (torque may not be zero)
Wire of mass m held up by B
BIL = mg
Quick check

A semicircular wire of radius R carrying I in uniform B ⊥ plane feels the same force as its diameter: F = BI(2R).

Biot–Savart law and standard fieldsMust

Biot–Savart
dB = (μ0/4π) I dl × r̂ / r2; μ0 = 4π × 10−7 T m A−1
Centre of circular loop
B = μ0 N I / 2R
On axis of loop, distance x
B = μ0 N I R2 / 2(R2 + x2)3/2; far away (x ≫ R): B ≈ μ0 N I R2 / 2x3
Arc subtending angle φ (radians) at centre
B = μ0 I φ / 4πR; semicircle: μ0I/4R; quarter: μ0I/8R
Finite straight wire
B = (μ0 I / 4πd)(sinα + sinβ), angles measured from the perpendicular
Infinite wire
B = μ0 I / 2πd
Semi-infinite wire, point on perpendicular at its end
B = μ0 I / 4πd
Centre of square loop, side a
B = 2√2 μ0 I / πa
Inside long thick wire (radius R)
B = μ0 I r / 2πR2 (r < R); maximum at surface; outside μ0I/2πr
Magnetic field around a straight wire and forces between parallel currentsI out of page → B anticlockwiseB = μ₀I / 2πrsame direction → attractopposite → repelF/L = μ₀ I₁ I₂ / 2πd
Field lines are circles around the wire (right-hand thumb rule). Parallel wires pull together for like currents and push apart for unlike currents.

Ampere (NCERT definition): the current which, flowing in two infinitely long straight parallel wires 1 m apart in vacuum, produces a force of 2 × 10−7 N per metre of length.

Ampère's circuital law, solenoid, toroidMust

Ampère's law
∮ B · dl = μ0 Ienclosed
Long solenoid (inside)
B = μ0 n I, n = N/l; with core: μ0 μr n I
At the end of a long solenoid
B = μ0 n I / 2
Outside ideal solenoid
B ≈ 0
Toroid
B = μ0 N I / 2πr inside the core; zero in the open space and outside
Remember

Inside a long solenoid B is uniform and independent of radius and of position along the length (away from the ends).

Torque on a loop and the magnetic dipoleMust

Rotating current loop with torque and magnetic momentBmCoil in uniform B (edge view)τ = m × B, |τ| = NIAB sinθU = −m·B = −mB cosθθ = angle between normal m and BStart: plane ∥ B, θ = 90°, torque maximumEnd: plane ⊥ B, θ = 0°, torque zero, stableForces on opposite sides: equal, opposite→ net force zero, only torque
Forces on the two long sides stay vertical and equal. As the coil turns, their lever arm shrinks, so torque falls to zero when the normal lines up with B.
Magnetic moment
m = N I A (direction by right-hand rule, along the normal)
Torque
τ = m × B = NIAB sinθ
Potential energy
U = −mB cosθ; minimum (stable) at θ = 0°, maximum (unstable) at 180°
Work to rotate from θ1 to θ2
W = mB (cosθ1 − cosθ2)
Revolving electron
μl = (e/2me) l; smallest value μB = eh/4πme = 9.27 × 10−24 A m2
Gyromagnetic ratio
μl / l = e/2me = 8.8 × 1010 C kg−1

Moving coil galvanometer, ammeter, voltmeterMust

Radial field (concave poles + soft iron core) keeps the plane of the coil parallel to B, so sinθ = 1 at every deflection: NIAB = kφ → φ = (NAB/k) I, a linear scale.

Current sensitivity
φ/I = NAB/k
Voltage sensitivity
φ/V = NAB/(kRG)
Double N (same wire type)
current sensitivity ×2, resistance ×2 → voltage sensitivity unchanged
Ammeter
shunt S = Ig G / (I − Ig) in parallel; RA = GS/(G+S) ≈ S, ideal 0
Voltmeter
R = V/Ig − G in series; RV = G + R, ideal ∞
Converting a galvanometer into an ammeter and a voltmeterGIg (tiny)S (small), carries I − IgAmmeter: S = Ig G / (I − Ig)shunt in parallel · joined in seriesGR (large)Voltmeter: R = V/Ig − Gresistance in series · joined in parallel
Most of the current bypasses G through the shunt. For a voltmeter the large series resistance keeps the drawn current tiny.

CyclotronGap content

Gap content

Not in the rationalised NCERT text but part of the NTA physics list for this chapter and a regular coaching question.

Cyclotron dees with spiral path of an accelerated ionD₁D₂beam outB out of page, AC across the gapfc = qB / 2πmKmax = q²B²R² / 2mResonance: oscillator f = fcRadius grows each half-turn,time per half-turn stays the sameNot for neutrons (no charge) orfast electrons (relativistic)
Each crossing of the gap adds energy; the orbit radius grows but the half-period qB/2πm stays constant, so a fixed-frequency oscillator stays in step.
Cyclotron frequency
fc = qB/2πm (angular: ω = qB/m)
Maximum K (dee radius R)
Kmax = q2 B2 R2 / 2m
Energy per revolution
2qV (two gap crossings)

Standard question patternsMust

Ratio of radii for p, d, α
First identify what is equal (V, K, p or v). Then use the matching proportionality from the table above.
Speed doubled in same B — effect on T
No change. r doubles, T = 2πm/qB unchanged.
Field at centre of combination of arcs and straight segments
Add μ0Iφ/4πR for arcs, μ0I/4πd (sinα + sinβ) for segments; segments pointing at the centre give zero.
Convert G to ammeter / voltmeter
S = IgG/(I − Ig) parallel; R = V/Ig − G series.
Torque on coil at angle
Use angle between normal and B. If 'plane makes α with B', θ = 90° − α.
Parallel wires force per length
F/L = 2 × 10−7 I1I2/d (SI); like currents attract.
Helix pitch
p = 2πm v cosθ/qB, and r uses v sinθ.

Traps that cost marksMust

Trap · Ratio reversal

r ∝ √m at fixed V: heavier particle → bigger radius, not smaller. For the same momentum only charge matters: α has the smaller radius.

Trap · Angle between plane and normal

'Plane of coil at 30° to B' means θ = 60° in NIAB sinθ. Using sin30° gives 0.50 instead of 0.87 — a ready-made wrong option.

Trap · f versus ω

Cyclotron f = qB/2πm ≈ 15.2 MHz for a proton at 1 T; qB/m ≈ 9.6 × 107 rad s−1 is ω. Options often include both.

Trap · Voltmeter series resistance

R = V/Ig − G. Forgetting to subtract G gives V/Ig, which will be sitting in the options.

Trap · Assertion–reason

'Magnetic force does no work' is true because force ⊥ velocity at every instant. Do not accept 'because B is uniform' as the reason.

Ten-question checkMust

Numericals are Python-verified. Aim for 9/10 in under 12 minutes.

Q1.A proton, a deuteron and an α-particle are accelerated through the same potential difference and enter a uniform magnetic field perpendicularly. The ratio of their radii rp : rd : rα is

  1. (A)1 : 2 : 2
  2. (B)1 : √2 : √2
  3. (C)1 : √2 : 1
  4. (D)√2 : 1 : 1
Show answer

Answer (B). r = √(2mV/q)/B ∝ √(m/q). Proton √(1/1) = 1, deuteron √(2/1) = √2, alpha √(4/2) = √2. 1 : √2 : 1 is the equal-K answer; 1 : 2 : 2 forgets the square root.

Q2.A circular coil of 100 turns and radius 10 cm carries 1 A. Magnetic field at its centre is

  1. (A)3.14 × 10−4 T
  2. (B)1.26 × 10−3 T
  3. (C)6.28 × 10−4 T
  4. (D)6.28 × 10−6 T
Show answer

Answer (C). B = μ0NI/2R = (4π × 10−7 × 100 × 1)/(2 × 0.1) = 6.28 × 10−4 T.

Q3.Two long parallel wires 5 cm apart carry 10 A and 5 A in opposite directions. The force per unit length between them is

  1. (A)2 × 10−4 N m−1, attractive
  2. (B)4 × 10−4 N m−1, repulsive
  3. (C)1 × 10−4 N m−1, attractive
  4. (D)2 × 10−4 N m−1, repulsive
Show answer

Answer (D). F/L = μ0I1I2/2πd = 2 × 10−7 × 10 × 5 / 0.05 = 2 × 10−4 N m−1. Opposite currents repel.

Q4.A galvanometer of resistance 50 Ω gives full-scale deflection for 5 mA. To use it as a voltmeter of range 0–10 V, one must connect

  1. (A)1950 Ω in series
  2. (B)2000 Ω in series
  3. (C)1950 Ω in parallel
  4. (D)2050 Ω in series
Show answer

Answer (A). R = V/Ig − G = 10/0.005 − 50 = 1950 Ω, in series. 2000 Ω forgets to subtract G.

Q5.A coil of 50 turns and area 0.02 m2 carries 2 A in a uniform field of 0.5 T. The plane of the coil makes 30° with the field. Torque on the coil is about

  1. (A)0.50 N m
  2. (B)0.87 N m
  3. (C)1.00 N m
  4. (D)0.43 N m
Show answer

Answer (B). Normal makes 90° − 30° = 60° with B. τ = NIAB sin60° = 50 × 2 × 0.02 × 0.5 × 0.866 = 0.87 N m.

Q6.A charged particle moves in a circle in a uniform magnetic field. If its speed is doubled, its time period of revolution

  1. (A)is doubled
  2. (B)remains the same
  3. (C)is halved
  4. (D)becomes four times
Show answer

Answer (B). T = 2πm/qB has no v. The radius doubles, the distance doubles, the time stays the same.

Q7.Assertion (A): The magnetic force on a moving charge does no work.
Reason (R): The magnetic force is always perpendicular to the velocity of the charge.

  1. (A)Both A and R are true, and R is the correct explanation of A
  2. (B)Both A and R are true, but R is not the correct explanation of A
  3. (C)A is true, but R is false
  4. (D)A is false, but R is true
Show answer

Answer (A). Power = F · v = 0 when F ⊥ v, so kinetic energy cannot change. R is the correct explanation.

Q8.A square loop of side a carries current I. The magnetic field at its centre is

  1. (A)√2 μ0 I / πa
  2. (B)μ0 I / √2 πa
  3. (C)4√2 μ0 I / πa
  4. (D)2√2 μ0 I / πa
Show answer

Answer (D). Each side: (μ0I/4π(a/2))(sin45° + sin45°) = √2μ0I/2πa. Four sides: 4 × √2μ0I/2πa = 2√2μ0I/πa.

Q9.A proton enters a uniform magnetic field with velocity at 60° to the field. The ratio of pitch of the helix to its radius is

  1. (A)2π/√3
  2. (B)2π√3
  3. (C)π/√3
  4. (D)2π
Show answer

Answer (A). p/r = [2πm v cos60°/qB]/[m v sin60°/qB] = 2π cot60° = 2π/√3 ≈ 3.63.

Q10.The cyclotron frequency of a proton in a magnetic field of 1 T is nearly (mp = 1.67 × 10−27 kg)

  1. (A)30.5 MHz
  2. (B)7.6 MHz
  3. (C)15.2 MHz
  4. (D)95.8 MHz
Show answer

Answer (C). f = qB/2πm = (1.6 × 10−19 × 1)/(2π × 1.67 × 10−27) ≈ 1.52 × 107 Hz. 95.8 × 106 is ω in rad s−1, not f.

60-second recap before the paperMust

  1. F = qvB sinθ; no work; r = mv/qB, T = 2πm/qB (independent of v).
  2. Same V: r ∝ √(m/q). Same K: r ∝ √m/q. Same p: r ∝ 1/q.
  3. Helix: r uses v sinθ, pitch uses v cosθ.
  4. Loop centre μ0NI/2R; arc μ0Iφ/4πR; infinite wire μ0I/2πd; semi-infinite μ0I/4πd.
  5. Solenoid μ0nI (end: half); toroid μ0NI/2πr; thick wire inside ∝ r.
  6. Parallel wires: F/L = μ0I1I2/2πd, like attract.
  7. τ = NIAB sinθ with θ from the normal; U = −mB cosθ; net force on closed loop in uniform B = 0.
  8. Ammeter: small shunt in parallel. Voltmeter: large R in series, R = V/Ig − G.
  9. Cyclotron: f = qB/2πm, Kmax = q2B2R2/2m.