Moving Charges and Magnetism — NEET Physics Mastery Pack

NCERT Class XII Physics, Chapter 4 · NEET Unit: Magnetic Effects of Current and Magnetism · Prepared for Aamirah

Part 1 — How to prepare so that nothing is missed

This chapter is not like Human Health and Disease. Nothing here is won by memory alone. NEET asks roughly 3–5 questions from this unit, and they split cleanly into two kinds:

Formula-substitution numericals (about 60%) — field of a wire/loop/solenoid, radius of a circular path, force between wires, shunt resistance. These are 30-second questions if the formula is automatic.
Direction and reasoning questions (about 40%) — right-hand rule, v × B direction, why the magnetic force does no work, why the field inside a toroid's hollow is zero, why current sensitivity ≠ voltage sensitivity. These are where marks are actually lost.

So the preparation has to attack two different things: formula reflex and vector direction sense. Learning the chapter by reading will not fix either one.

Syllabus alert — the cyclotron. The cyclotron was removed from the rationalised NCERT textbook, but it is listed in the NTA/NMC NEET syllabus for this unit. It is one of a handful of topics the NEET syllabus keeps even though NCERT dropped them. Do not skip it just because it is missing from the book — learn f = qB/2πm and K_max = q²B²R²/2m, and know that a cyclotron cannot accelerate neutrons (uncharged) or electrons (relativistic mass gain destroys resonance). Questions 24 and 25 in the test below cover it.

The 8-day mastery cycle

Day	Task	What "done properly" looks like
Day 0	Cold diagnostic	Take the 45-question test below untimed and without notes. Mark it. Sort every wrong answer into one of three buckets: (i) didn't know the formula, (ii) knew the formula but got the direction/sign wrong, (iii) arithmetic or unit slip. The bucket distribution tells you which of the days below matters most.
Day 1	NCERT + derivations	Read the chapter and derive, on paper: field of a straight wire from Biot–Savart, field at the centre and on the axis of a loop, field of a solenoid and a toroid from Ampère's law, force between two parallel wires, torque on a current loop. NEET does not ask for derivations, but doing them is what makes the formula stick and stops her mixing up `2R` with `2πR`.
Day 2	Formula sheet from memory	Close the book and write out Part 2 below on blank paper. Correct in red. Repeat until it comes out clean twice in a row. Every formula must come with its conditions — "μ₀nI" is only true inside a long solenoid, far from the ends.
Day 3	Direction drill	The highest-yield hour of the whole cycle. Take 40 sketches — wire, loop, solenoid, charge in a field, wire in a field — and mark the direction of B or F in each, out loud, in under 5 seconds each. Use the right-hand thumb rule for fields from currents and F = qv × B for forces. Remember to flip the answer for an electron.
Day 4	Charged-particle motion	Master the four cases as one family: straight line (v ∥ B), circle (v ⊥ B), helix (at an angle), undeviated (crossed E and B). Then drill the four versions of the radius formula: `r = mv/qB`, `r = p/qB`, `r = √(2mK)/qB`, `r = √(2mV/q)/B`. Almost every NEET question on this chapter that involves protons vs alpha particles is testing which of these four to use.
Day 5	Devices day	Galvanometer → ammeter (shunt in parallel) and → voltmeter (resistance in series). Do 15 conversion numericals until the two formulas never get swapped. Also nail the current-sensitivity vs voltage-sensitivity trap.
Day 6	PYQ drill	Ten years of previous-year questions from this unit in one sitting. For every error, write one line: the formula that was needed and the exact reason it was missed.
Day 7	Timed retest	Retake this same test, 45 minutes, strict. That is one minute per question, the real NEET pace. Target 42+/45. If speed is the problem rather than knowledge, the fix is more Day 2, not more Day 1.
Day 8	Compress	One A4 sheet in her own handwriting: every formula with its condition, plus the six direction rules. That sheet replaces the textbook for all future revision.

Spaced revision

Revise the one-page sheet on Day 11, Day 25, Day 55, Day 120, then fortnightly in the final three months. Ten minutes each. Re-take the test once a month — the goal is a repeatable 43–45, not a one-off good day.

The six traps that actually cost marks in this chapter

2R vs 2πR. Centre of a loop is μ₀I/2R; a straight wire is μ₀I/2πR. Mixing these is the single most common error in the chapter.
Forgetting that the magnetic force does no work. Speed and kinetic energy of a charged particle in a pure magnetic field never change — only direction does.
Period vs radius. T = 2πm/qB is independent of speed and radius. Faster particles simply trace bigger circles in the same time. This is the whole principle of the cyclotron.
Net force vs net torque on a loop. In a uniform field the net force on any closed current loop is zero, but the torque generally is not.
Current sensitivity vs voltage sensitivity. Adding turns raises current sensitivity, but it also raises the coil resistance — so voltage sensitivity need not increase.
Sign flip for the electron. Work out the direction for a positive charge first, then reverse it. Doing it in one step is how careless marks are lost.

Part 2 — Formula and concept sheet

2.1 Magnetic field produced by currents

Situation	Formula	Conditions / notes
Biot–Savart law	`dB = (μ₀/4π)·(I dl sinθ)/r²`	Vector form `dB = (μ₀/4π) I dl×r̂/r²`. Zero at points lying along the direction of the element (θ = 0).
Infinite straight wire	`B = μ₀I/2πa`	Field lines are concentric circles; direction by right-hand thumb rule. `μ₀/4π = 10⁻⁷ T m A⁻¹`.
Finite straight wire	`B = (μ₀I/4πa)(sinθ₁ + sinθ₂)`	θ measured from the perpendicular to the two ends.
Centre of a circular coil	`B = μ₀NI/2R`	N turns. Note 2R, not 2πR.
Arc subtending angle φ (radians)	`B = μ₀Iφ/4πR`	Semicircle (φ = π) gives `μ₀I/4R`.
Axis of a circular loop	`B = μ₀NIR²/2(R²+x²)^3/2`	For x ≫ R this becomes `B = μ₀·2m/4πx³` — the loop acts as a magnetic dipole.
Centre of a square loop of side a	`B = 2√2 μ₀I/πa`	Derived by adding four finite wires.
Ampère's circuital law	`∮B·dl = μ₀I_enclosed`	The line integral depends only on the enclosed current — not on the shape or size of the loop, nor on currents outside it.
Long solenoid (inside)	`B = μ₀nI`	n = turns per unit length. Uniform inside, nearly zero outside. At either end the field is `μ₀nI/2`.
Toroid	`B = μ₀NI/2πr`	N = total turns. Field is zero both outside the toroid and in the empty space enclosed by it.
Inside a thick cylindrical conductor	`B = μ₀Ir/2πR²` for r < R	Field rises linearly inside, falls as 1/r outside.

2.2 Motion of a charged particle

Quantity	Formula	Notes
Lorentz force	`F = q(E + v×B)`	Magnetic part is always ⊥ to v, so it does no work; speed and KE are constant.
Radius of circular path	`r = mv/qB = p/qB = √(2mK)/qB`	Accelerated through potential V: `r = (1/B)√(2mV/q)`.
Period, frequency, angular frequency	`T = 2πm/qB`, `f = qB/2πm`, `ω = qB/m`	All independent of speed and radius.
Helical path	pitch `= v cosθ · T = 2πmv cosθ/qB`	Radius uses only the perpendicular component `v sinθ`.
Velocity selector (crossed E, B)	`v = E/B`	Particle passes undeviated regardless of its charge and mass.
Cyclotron	`f = qB/2πm`, `K_max = q²B²R²/2m`	In the NEET syllabus though dropped from NCERT. Cannot accelerate neutrons (no charge) or electrons (relativistic mass increase breaks resonance).

2.3 Forces, torques and dipoles

Quantity	Formula	Notes
Force on a current-carrying conductor	`F = BIL sinθ`, `F = IL×B`	Net force on any closed loop in a uniform field is zero.
Force between parallel wires	`F/l = μ₀I₁I₂/2πd`	Same direction → attract; opposite → repel.
Definition of the ampere	`2 × 10⁻⁷ N m⁻¹`	Force per metre between two infinitely long parallel conductors 1 m apart each carrying 1 A.
Magnetic moment of a loop	`m = NIA`	Direction by right-hand rule, normal to the loop plane. Unit: A m².
Torque on a loop	`τ = NIAB sinθ`, `τ = m×B`	θ is between m and B. Torque is maximum when the plane of the coil is parallel to B, and zero when the plane is perpendicular to B.
Potential energy	`U = −m·B = −mB cosθ`	Stable equilibrium at θ = 0, unstable at θ = 180°.
Revolving electron	`μ_l = evr/2 = (e/2m)L`	Gyromagnetic ratio `e/2m = 8.8 × 10¹⁰ C kg⁻¹`. Bohr magneton `μ_B = eh/4πm = 9.27 × 10⁻²⁴ A m²`.

2.4 Moving coil galvanometer

At equilibrium NIAB = kφ, so I = (k/NAB)φ — deflection is proportional to current.
The radial magnetic field (concave pole pieces + soft-iron core) keeps the plane of the coil always parallel to B, so sinθ = 1 always and the scale is linear.
Current sensitivity = φ/I = NAB/k · Voltage sensitivity = φ/V = NAB/kR. Increasing N raises current sensitivity but also raises R, so voltage sensitivity need not improve.
Ammeter: small shunt in parallel, S = I_gG/(I − I_g). Connected in series; ideal resistance is zero.
Voltmeter: large resistance in series, R = V/I_g − G. Connected in parallel; ideal resistance is infinite.

Part 3 — Full Chapter Test

Subject: Physics — Moving Charges and Magnetism | Questions: 45 | Maximum marks: 180
Time: 45 minutes (NEET pace) | Marking: +4 correct, −1 incorrect, 0 unattempted
Constants: μ₀ = 4π × 10⁻⁷ T m A⁻¹, μ₀/4π = 10⁻⁷ T m A⁻¹, e = 1.6 × 10⁻¹⁹ C, m_p = 1.67 × 10⁻²⁷ kg
On the first attempt, do it untimed. Only from the second attempt onwards should the 45-minute limit be enforced.

1. A long straight wire carries a current of 5 A. The magnetic field at a point 2 cm from the wire is:

(a) 2.5 × 10⁻⁵ T

(b) 5 × 10⁻⁵ T

(d) 5 × 10⁻⁶ T

2. According to the Biot–Savart law, the magnetic field due to a current element is zero at points lying:

(a) On the perpendicular bisector of the element

(b) Along the axis of the element, i.e. along dl

(d) Nowhere; it is never zero

3. A circular coil of 100 turns and radius 10 cm carries a current of 1 A. The magnetic field at its centre is approximately:

(a) 6.28 × 10⁻⁴ T

(b) 6.28 × 10⁻⁵ T

(d) 1.26 × 10⁻³ T

4. A semicircular wire of radius 0.2 m carries a current of 4 A. The magnetic field at the centre of the semicircle is:

(a) 1.26 × 10⁻⁵ T

(b) 6.28 × 10⁻⁶ T

(d) 2.51 × 10⁻⁵ T

5. A wire of fixed length carrying a constant current is bent into a single circular turn, giving a field B at its centre. If the same wire is rewound into two concentric turns of half the radius, the field at the centre becomes:

(a) 2B

(b) 4B

(d) B/4

6. The line integral ∮B·dl around a closed path depends on:

(a) The shape and size of the path

(b) Only the currents enclosed by the path

(d) The magnetic field at every point on the path only

7. A long solenoid has 500 turns per metre and carries a current of 2 A. The magnetic field well inside the solenoid is:

(a) 1.26 × 10⁻³ T

(b) 6.28 × 10⁻⁴ T

(d) 4.00 × 10⁻⁴ T

8. A toroid of mean radius 0.1 m has 500 turns and carries a current of 2 A. The magnetic field inside the core is:

(a) 1 × 10⁻³ T

(b) 2 × 10⁻³ T

(d) 2 × 10⁻⁴ T

9. The magnetic field at the centre of a square loop of side a carrying a current I is:

(a) μ₀I/2a

(b) 2√2 μ₀I/πa

(d) μ₀I/4πa

10. Two long parallel wires 10 cm apart each carry 10 A in opposite directions. The magnitude of the magnetic field at a point midway between them is:

(a) Zero

(b) 4 × 10⁻⁵ T

(d) 2 × 10⁻⁵ T

11. If the two wires in the previous question carried their currents in the same direction, the field midway between them would be:

(a) Zero

(b) 4 × 10⁻⁵ T

(d) 1.6 × 10⁻⁴ T

12. At a point on the axis of a small current loop, far from the loop (x ≫ R), the magnetic field varies with distance as:

(a) 1/x

(b) 1/x²

(d) 1/x⁴

13. For a circular loop of radius R, the ratio of the magnetic field at the centre to that at an axial point a distance R from the centre is:

(a) 2√2 : 1

(b) √2 : 1

(d) 1 : 2√2

14. Inside a long straight thick cylindrical conductor of radius R carrying a uniformly distributed current, the magnetic field at a distance r < R from the axis:

(a) Is zero everywhere

(b) Increases linearly with r

(d) Decreases as 1/r²

15. The magnetic field at either end of a long solenoid, compared with the field at its centre, is:

(a) Equal

(b) Half

(d) Zero

16. A charged particle enters a uniform magnetic field with its velocity parallel to the field. Its path is:

(a) A circle

(b) A helix

(d) A parabola

17. A magnetic field does no work on a moving charged particle because:

(a) The field is conservative

(b) The force is always perpendicular to the velocity

(d) The particle moves in a closed path

18. A proton and an alpha particle having the same kinetic energy enter a uniform magnetic field perpendicular to it. The ratio of the radii of their circular paths (r_p : r_α) is:

(a) 1 : 1

(b) 1 : 2

(d) 1 : √2

19. The time period of a charged particle moving in a circular path in a uniform magnetic field is:

(a) Proportional to its speed

(b) Proportional to the radius of the path

(d) Inversely proportional to its speed

20. A proton and an alpha particle are accelerated through the same potential difference and then enter a uniform magnetic field perpendicular to their velocities. The ratio r_p : r_α is:

(a) 1 : 1

(b) 1 : √2

(d) 1 : 2

21. In a velocity selector, an electric field of 2 × 10⁴ V m⁻¹ is applied perpendicular to a magnetic field of 0.1 T. The speed of the particles that pass through undeviated is:

(a) 2 × 10³ m s⁻¹

(b) 2 × 10⁵ m s⁻¹

(d) 5 × 10⁴ m s⁻¹

22. A charged particle enters a uniform magnetic field with its velocity making an angle θ (0 < θ < 90°) with the field. The pitch of its helical path is:

(a) 2πmv sinθ/qB

(b) 2πmv cosθ/qB

(d) mv sinθ/qB

23. The cyclotron frequency of a proton in a magnetic field of 0.5 T is approximately:

(a) 7.6 × 10⁶ Hz

(b) 7.6 × 10⁷ Hz

(d) 3.8 × 10⁶ Hz

24. A cyclotron cannot be used to accelerate:

(a) Protons

(b) Alpha particles

(d) Deuterons

25. The maximum kinetic energy of a particle of charge q and mass m emerging from a cyclotron of dee radius R operating at field B is:

(a) q²B²R²/2m

(b) qBR/2m

(d) qB²R²/2m

26. An electron moves horizontally towards the east in a magnetic field directed vertically downwards. The magnetic force on the electron is directed towards the:

(a) North

(b) South

(d) Vertically upwards

27. The work done by the magnetic force on a charged particle completing one full circular revolution in a uniform magnetic field is:

(a) Zero

(b) qvB × 2πr

(d) Half its kinetic energy

28. A proton and an alpha particle move in circular paths in the same uniform magnetic field. The ratio of their time periods T_p : T_α is:

(a) 1 : 1

(b) 1 : 2

(d) 1 : 4

29. A straight wire of length 0.5 m carrying 4 A is placed in a uniform magnetic field of 0.2 T such that the wire makes an angle of 30° with the field. The force on the wire is:

(a) 0.4 N

(b) 0.2 N

(d) 0.35 N

30. The net force on a closed current loop of arbitrary shape placed in a uniform magnetic field is:

(a) Always zero

(b) Always maximum

(d) Equal to BIL for any loop

31. Two long parallel wires 10 cm apart each carry a current of 5 A in the same direction. The force per unit length between them is:

(a) 5 × 10⁻⁵ N m⁻¹, attractive

(b) 5 × 10⁻⁵ N m⁻¹, repulsive

(d) 1 × 10⁻⁴ N m⁻¹, repulsive

32. One ampere is that steady current which, when maintained in two infinitely long parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force per unit length of:

(a) 2 × 10⁻⁷ N m⁻¹

(b) 1 × 10⁻⁷ N m⁻¹

(d) 2 × 10⁻⁵ N m⁻¹

33. A coil of 100 turns and area 0.01 m² carries 2 A and is placed in a uniform field of 0.5 T with the plane of the coil parallel to the field. The torque on the coil is:

(a) Zero

(b) 0.5 N m

(d) 2 N m

34. A circular coil of 50 turns and radius 5 cm carries a current of 2 A. Its magnetic dipole moment is approximately:

(a) 0.39 A m²

(b) 0.79 A m²

(d) 7.85 A m²

35. A current loop placed in a uniform magnetic field is in stable equilibrium when its magnetic moment m is:

(a) Parallel to B

(b) Antiparallel to B

(d) At 45° to B

36. A wire of given length carrying a current is bent first into a circle and then into a square. The magnetic moment is:

(a) Greater for the circle

(b) Greater for the square

(d) Zero in both cases

37. For an electron revolving in a circular orbit, the ratio of its orbital magnetic moment to its orbital angular momentum is:

(a) e/m

(b) e/2m

(d) m/2e

38. The value of the Bohr magneton is approximately:

(a) 9.27 × 10⁻²⁴ A m²

(b) 9.27 × 10⁻²¹ A m²

(d) 8.8 × 10¹⁰ A m²

39. The current sensitivity of a moving coil galvanometer is given by:

(a) NAB/k

(b) NAB/kR

(d) kR/NAB

40. Increasing the number of turns of the coil of a moving coil galvanometer:

(a) Increases current sensitivity but need not increase voltage sensitivity

(b) Increases both sensitivities in the same proportion

(d) Has no effect on either sensitivity

41. A galvanometer of resistance 100 Ω gives full-scale deflection for 1 mA. To convert it into an ammeter of range 1 A, the shunt required is approximately:

(a) 0.1 Ω, in parallel

(b) 0.1 Ω, in series

(d) 999 Ω, in series

42. A galvanometer of resistance 50 Ω gives full-scale deflection for 2 mA. To convert it into a voltmeter reading up to 10 V, the resistance required is:

(a) 4950 Ω in series

(b) 5050 Ω in series

(d) 5000 Ω in parallel

43. An ideal ammeter and an ideal voltmeter have, respectively:

(a) Zero and infinite resistance

(b) Infinite and zero resistance

(d) Infinite and infinite resistance

44. A radial magnetic field is used in a moving coil galvanometer so that:

(a) The plane of the coil is always parallel to the field, giving a linear scale

(b) The coil rotates faster

(d) The magnetic field strength increases with deflection

45. Assertion (A): The magnetic field is zero both outside a toroid and in the empty space enclosed by it.
Reason (R): For an Amperian loop drawn in either of these regions, the net current enclosed is zero.

(a) Both A and R are true and R correctly explains A

(b) Both A and R are true but R does not explain A

(d) A is false, R is true

Part 4 — Answer key with worked solutions (open only after attempting)

Q	Ans	Working / reason
1	b	B = μ₀I/2πa = (2×10⁻⁷ × 5)/0.02 = 5 × 10⁻⁵ T.
2	b	dB ∝ sinθ, and θ = 0 for points along the element, so dB = 0 there.
3	a	B = μ₀NI/2R = (4π×10⁻⁷ × 100 × 1)/(2 × 0.1) = 2π × 10⁻⁴ ≈ 6.28 × 10⁻⁴ T.
4	b	Semicircle: B = μ₀I/4R = (4π×10⁻⁷ × 4)/(4 × 0.2) = 2π × 10⁻⁶ ≈ 6.28 × 10⁻⁶ T.
5	b	Length fixed: 2πR = 2(2πR′) so R′ = R/2 with N = 2. B′ = μ₀(2)I/2(R/2) = 4 × μ₀I/2R = 4B. In general B ∝ N².
6	b	Ampère's law: ∮B·dl = μ₀I_enclosed. External currents contribute to B at points on the loop but not to the integral.
7	a	B = μ₀nI = 4π×10⁻⁷ × 500 × 2 = 4π × 10⁻⁴ ≈ 1.26 × 10⁻³ T.
8	b	B = μ₀NI/2πr = (2×10⁻⁷ × 500 × 2)/0.1 = 2 × 10⁻³ T.
9	b	Each side is a finite wire at distance a/2 with θ₁ = θ₂ = 45°: B_side = (μ₀I/4π(a/2))(√2) = √2μ₀I/2πa. Four sides give 2√2 μ₀I/πa.
10	c	Each wire gives 2×10⁻⁷ × 10/0.05 = 4 × 10⁻⁵ T at the midpoint. For antiparallel currents the two fields point the same way and add: 8 × 10⁻⁵ T.
11	a	For parallel currents the two fields are opposite at the midpoint and cancel exactly.
12	c	Far field of a magnetic dipole: B = μ₀·2m/4πx³.
13	a	B_centre = μ₀I/2R. At x = R, B = μ₀IR²/2(2R²)^3/2 = μ₀I/(4√2 R). Ratio = 2√2 : 1.
14	b	Enclosed current ∝ r², so B = μ₀Ir/2πR² — linear in r inside, then 1/r outside.
15	b	At an end the solenoid contributes only "half" its length, giving B = μ₀nI/2.
16	c	v ∥ B ⇒ v × B = 0 ⇒ no magnetic force ⇒ undeflected straight line.
17	b	F ⊥ v at every instant, so F·ds = 0 and no work is done. Speed and KE stay constant.
18	a	r = √(2mK)/qB. r_p/r_α = (√m_p/q_p)·(q_α/√m_α) = (√1 × 2)/(1 × √4) = 1. The radii are equal — a favourite NEET trap.
19	c	T = 2πm/qB contains neither v nor r. This is why a cyclotron works at a fixed frequency.
20	b	Here r = (1/B)√(2mV/q) ∝ √(m/q). Proton: √(1/1) = 1; alpha: √(4/2) = √2. So 1 : √2. Contrast carefully with Q18.
21	b	qE = qvB ⇒ v = E/B = 2×10⁴/0.1 = 2 × 10⁵ m s⁻¹.
22	b	Pitch = (component along B) × (period) = v cosθ × 2πm/qB.
23	a	f = qB/2πm = (1.6×10⁻¹⁹ × 0.5)/(2π × 1.67×10⁻²⁷) ≈ 7.6 × 10⁶ Hz.
24	c	A neutron is uncharged, so it experiences no electric or magnetic force. (Electrons are also impractical because of relativistic mass increase, but the neutron is the outright impossibility.)
25	a	At the dee edge v = qBR/m, so K = ½mv² = q²B²R²/2m.
26	b	Take east = +x, north = +y, up = +z; B = −Bẑ. For a positive charge, v×B = vx̂ × (−Bẑ) = +vBŷ (north). The electron's charge is negative, so the force reverses to south.
27	a	The magnetic force never does work, so it is zero over any path, closed or not.
28	b	T ∝ m/q. Proton: 1/1 = 1. Alpha: 4/2 = 2. So T_p : T_α = 1 : 2.
29	b	F = BIL sinθ = 0.2 × 4 × 0.5 × sin30° = 0.2 N.
30	a	F = I(∮dl) × B, and ∮dl = 0 for any closed loop. Note the torque is generally not zero.
31	a	F/l = μ₀I₁I₂/2πd = 2×10⁻⁷ × 25/0.1 = 5 × 10⁻⁵ N m⁻¹. Same direction ⇒ attraction.
32	a	Substituting I₁ = I₂ = 1 A and d = 1 m gives 2 × 10⁻⁷ N m⁻¹ — the defining value.
33	c	Plane parallel to B means m ⊥ B, so sinθ = 1 and torque is maximum: τ = NIAB = 100 × 2 × 0.01 × 0.5 = 1 N m.
34	b	m = NIA = 50 × 2 × π(0.05)² = 100 × 7.854×10⁻³ ≈ 0.785 A m².
35	a	U = −mB cosθ is minimum at θ = 0, i.e. m aligned with B.
36	a	For perimeter L: circle area = L²/4π ≈ 0.0796L²; square area = L²/16 = 0.0625L². Since m = IA, the circle wins. For a fixed perimeter the circle always encloses the largest area.
37	b	μ_l = evr/2 and L = mvr, so μ_l/L = e/2m — the gyromagnetic ratio, 8.8 × 10¹⁰ C kg⁻¹.
38	a	μ_B = eh/4πm ≈ 9.27 × 10⁻²⁴ A m². Option (d) is the gyromagnetic ratio, a common distractor.
39	a	φ/I = NAB/k. Option (b) is voltage sensitivity.
40	a	Current sensitivity ∝ N, but the coil resistance R also rises with N, and voltage sensitivity = NAB/kR — so it may not improve at all.
41	a	S = I_gG/(I − I_g) = (10⁻³ × 100)/(1 − 10⁻³) ≈ 0.1 Ω, connected in parallel.
42	a	R = V/I_g − G = 10/(2×10⁻³) − 50 = 5000 − 50 = 4950 Ω, connected in series.
43	a	An ammeter goes in series so it must not drop any voltage (R = 0); a voltmeter goes in parallel so it must draw no current (R = ∞).
44	a	Concave pole pieces and a soft-iron core make B always lie in the plane of the coil, so sinθ = 1 for every deflection and φ ∝ I exactly.
45	a	Both regions enclose zero net current, so ∮B·dl = 0; with the symmetry of the toroid this forces B = 0. R is the correct explanation.

Reading the score

Marks / 180	Correct	What it means and what to do
168–180	43–45	Exam-ready. Move to spaced revision only.
130–167	34–42	Formulas are in place, application is shaky. Repeat Day 3 (directions) and Day 4 (charged-particle motion).
90–129	24–33	Formula recall is unreliable. Repeat Days 1–2 and rewrite the formula sheet from memory until it is clean.
Below 90	Under 24	Run the full 8-day cycle from Day 1 without shortcuts. Do not attempt timed papers yet.

Diagnose by bucket, not by score. Group the wrong answers: mostly Q1–15 means the field formulas aren't automatic; mostly Q16–28 means charged-particle motion needs rebuilding; mostly Q39–44 means the galvanometer section was skimmed. Fix the block, not the chapter.