NEET Physics · Class 12 · Chapter 4 · All sections
Moving Charges & Magnetism Full Chapter Mixed Test
The missing piece. In the five topic packs every question sits under a heading that gives away the formula. Here nothing is labelled — exactly like the real paper.
Not new material — a way of seeing how the ten sections fit together, so that an unlabelled question stops feeling like a guessing game.
1. The chapter has exactly two halves
Everything in Chapter 4 answers one of two questions. Deciding which one a question is asking is the first and most important step.
Half A — What does a magnetic field DO?
Half B — Where does a magnetic field COME FROM?
A field already exists. Something is placed in it and gets pushed or turned.
4.2.2 Force on a moving charge 4.3 The circle and the helix 4.2.3 Force on a current-carrying wire 4.9 Torque on a loop 4.10 The galvanometer (torque put to work)
A current exists. We want to know the field it creates.
4.4 Biot–Savart law (one crumb of wire) 4.5 Circular coil 4.6 Ampere's law, straight wire, thick wire 4.7 Solenoid
Answers come out in: newtons, newton metres, or degrees of deflection
Answers come out in: tesla
Section 4.8 is the join. Two parallel wires: one wire makes a field (Half B), and the other wire feels it (Half A). That is why it comes last — it needs both halves.
2. The unit tells you the half
Look at the options before you look at the numbers.
Answers in tesla → Half B. They want a field. Ask: wire, coil, arc, or solenoid?
Answers in newton → Half A. Force on a charge or on a wire.
Answers in newton per metre → the two-wire formula, almost always.
Answers in N m → torque on a loop.
Answers in A m² → magnetic moment.
Answers in ohm → galvanometer conversion.
Answers in metre or cm → radius of a circular path, r = mv/qB.
Answers in second or Hz → time period or frequency, T = 2πm/qB.
This single glance eliminates most of the chapter before any arithmetic begins.
3. The keyword decoder
Once you know the half, one phrase in the question usually names the formula outright.
If the question says…
It wants
"long straight wire", "infinite conductor"
B = μ₀I / 2πr
"circular coil of N turns", "at the centre"
B = μ₀NI / 2R
"semicircular", "arc", "bent wire"
B = μ₀Iθ / 4πR
"tightly wound", "turns per unit length", with a length given
B = μ₀nI
"solid cylinder", "uniformly distributed", "inside the wire"
B = μ₀Ir / 2πa²
"moving with speed v", "enters perpendicular"
F = qvB, r = mv/qB
"suspended in mid-air", "does not slide", "floats"
BIL = mg
"two parallel conductors", "force per unit length"
f = μ₀I₁I₂ / 2πd
"magnetic moment", "dipole moment"
m = NIA
"torque", "plane of the coil makes an angle"
τ = mB sin θ
"full-scale deflection", "shunt", "range"
S = I_gG/(I−I_g) or R = V/I_g − G
4. The seven traps, collected
Across the whole chapter these are the mistakes that actually cost marks. Every one of them appears somewhere in the 50 questions below.
θ from the normal, not the plane. In τ = mB sin θ, "plane parallel to B" means θ = 90° and maximum torque.
Coil versus solenoid. N is total turns and R matters; n is turns per metre and the radius is irrelevant.
1/r, not 1/r². A straight wire's field and the two-wire force both fall off as the first power of distance.
T does not depend on speed. There is no v in T = 2πm/qB.
Like currents attract — the opposite of the rule for charges.
The right-hand rules are two different rules. Straight wire: thumb = current. Loop: thumb = field.
Unit conversions. cm to m before squaring; grams to kilograms; mA to A; turns per mm to turns per metre.
5. How to sit this test
Time it. 50 questions in 60 minutes, in one sitting, without the formula sheet in front of her.
Mark every question she was unsure about, even the ones she got right. A lucky guess is a gap.
After marking, sort the errors by the section tag shown in each solution. That tells you which of the five topic packs to go back to — rather than guessing from a general priority order.
Two or more errors in the same section means redo that pack. One isolated error usually means a slip, not a gap.
Part 2 — Master formula sheet
Every formula in Chapter 4, grouped by the two halves. This is the one page to revise from the night before.
Half A — What a field does
Quantity
Formula
Note
Lorentz force
F = q(E + v × B)
Electric part + magnetic part.
Magnetic force on a charge
F = q v B sin θ
Zero if v = 0 or v ∥ B.
Work done by a magnetic force
W = 0
Speed and KE never change.
Radius of the circular path
r = mv/qB = p/qB = √(2mK)/qB
Three forms — check what is fixed.
Time period and frequency
T = 2πm/qB, ν = qB/2πm
Independent of speed.
Pitch of the helix
p = 2πmv cos θ / qB
Uses the along-field component.
Velocity selector
v = E / B
Undeflected in crossed fields.
Force on a wire
F = B I L sin θ
B must be external. L_eff for bent wires.
Wire in equilibrium
B I L = m g
Incline: I = (m/L)g tan α / B.
Net force on a closed loop
F = 0
Only in a uniform field.
Magnetic moment
m = N I A
Unit A m².
Torque on a loop
τ = m B sin θ
θ measured from the normal.
Energy and work
U = −mB cos θ, W = mB(cos θ₁ − cos θ₂)
Full flip: W = 2mB.
Galvanometer
N I A B = k θ
θ ∝ I ⇒ linear scale.
Sensitivities
I_s = NAB/k, V_s = NAB/kG
More turns raises I_s, not necessarily V_s.
Ammeter / voltmeter
S = I_gG/(I−I_g), R = V/I_g − G
Shunt parallel and small; R series and large.
Half B — Where a field comes from
Configuration
Field
Note
Current element
dB = (μ₀/4π) I dl sin θ / r²
Zero along the element's own line.
Constants
μ₀/4π = 10⁻⁷, μ₀/2π = 2 × 10⁻⁷
ε₀μ₀ = 1/c².
Long straight wire
B = μ₀I / 2πr
Falls as 1/r.
Thick wire, inside / outside
μ₀Ir/2πa² / μ₀I/2πr
Rises to the surface, then falls.
Hollow pipe, inside
B = 0
No enclosed current.
Ampere's law
∮B·dl = μ₀I_enclosed
Zero integral ≠ zero field.
Coil centre / axis
μ₀NI/2R / μ₀NIR²/2(x²+R²)^{3/2}
Far away, falls as 1/x³.
Arc / semicircle / quarter
μ₀Iθ/4πR / μ₀I/4R / μ₀I/8R
θ in radians.
Solenoid, inside / end
μ₀nI / μ₀nI/2
n = N/L. Radius irrelevant.
Two parallel wires
f = μ₀I₁I₂ / 2πd
Parallel attract, antiparallel repel.
Definition of the ampere
f = 2 × 10⁻⁷ N/m
1 A each, 1 m apart, in vacuum.
Part 3 — 50 mixed questions with step-by-step solutions
Nothing is labelled by topic. Work out which formula each question wants before you calculate anything — that is the skill being tested. Each solution reveals the section it came from, so errors can be traced back to the right pack.
Q1Mixed
A proton moving at 2 × 10⁶ m/s enters a magnetic field of 0.5 T at right angles to it. The force on the proton is:
(a) 8 × 10⁻¹⁴ N
(b) 1.6 × 10⁻¹³ N
(c) 3.2 × 10⁻¹³ N
(d) 1.6 × 10⁻¹⁴ N
Show step-by-step solution
Givenq = 1.6 × 10⁻¹⁹ C, v = 2 × 10⁶ m/s, B = 0.5 T, θ = 90°
AskedForce F
ConceptAnswer wanted in newtons and a moving charge is named ⇒ magnetic force on a charge.4.2.2 Lorentz force
FormulaF = q v B sin θ
Solutionsin 90° = 1 F = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.5 = 1.6 × 10⁻¹³ N
Answer: 1.6 × 10⁻¹³ N
Q2Mixed
A circular coil of 200 turns and radius 20 cm carries a current of 2 A. The field at its centre is:
(a) 6.28 × 10⁻⁴ T
(b) 1.256 × 10⁻³ T
(c) 2.51 × 10⁻³ T
(d) 3.14 × 10⁻⁴ T
Show step-by-step solution
GivenN = 200, R = 20 cm = 0.2 m, I = 2 A
AskedField at the centre
ConceptAnswer in tesla, and the words 'circular coil' with 'at the centre' name the formula.4.5 Circular coil
A charged particle moving perpendicular to a magnetic field has time period T. If its speed is halved, the time period becomes:
(a) T/2
(b) T
(c) 2T
(d) 4T
Show step-by-step solution
GivenSpeed halved, same q, m and B
AskedNew time period
ConceptCheck whether v appears in the formula at all — it does not.4.3 Circular motion
FormulaT = 2π m / q B
SolutionT depends only on m, q and B. Halving the speed halves the radius, so the smaller circle is covered at the lower speed. The two effects cancel and T is unchanged.
Answer: T
Q6Mixed
Two long parallel wires 12 cm apart carry currents of 3 A and 4 A. The force per unit length between them is:
(a) 1 × 10⁻⁵ N/m
(b) 2 × 10⁻⁵ N/m
(c) 4 × 10⁻⁵ N/m
(d) 2 × 10⁻⁶ N/m
Show step-by-step solution
GivenI₁ = 3 A, I₂ = 4 A, d = 12 cm = 0.12 m
AskedForce per unit length f
ConceptAnswer in newtons per metre with two currents given ⇒ the two-wire formula.4.8 Parallel wires
A wire bent into a quarter circle of radius 10 cm carries 4 A. The field at the centre is:
(a) 3.14 × 10⁻⁶ T
(b) 6.28 × 10⁻⁶ T
(c) 1.256 × 10⁻⁵ T
(d) 2.51 × 10⁻⁵ T
Show step-by-step solution
Givenθ = 90° = π/2 rad, R = 0.1 m, I = 4 A
AskedField at the centre
Concept'Quarter circle' ⇒ arc formula. Convert the angle to radians first.4.5 Arc
FormulaB = μ₀ I θ / 4πR = 10⁻⁷ I θ / R
Solutionθ = π/2 = 1.5708 rad B = 10⁻⁷ × 4 × 1.5708 / 0.1 = 10⁻⁷ × 62.83 = 6.28 × 10⁻⁶ T
Answer: 6.28 × 10⁻⁶ T
Q13Mixed
A horizontal wire of length 1 m and mass 100 g is held in mid-air by a perpendicular horizontal field of 0.5 T. The current in it is:
(a) 0.98 A
(b) 1.96 A
(c) 2.45 A
(d) 4.9 A
Show step-by-step solution
GivenL = 1 m, m = 100 g = 0.1 kg, B = 0.5 T, g = 9.8 m/s²
AskedCurrent I
ConceptThe words 'held in mid-air' signal the balance family: magnetic force equals weight.4.2.3 Balance
FormulaB I L = m g ⇒ I = mg / BL
Solutionm = 100 g = 0.1 kg I = (0.1 × 9.8) / (0.5 × 1) = 0.98 / 0.5 = 1.96 A
Answer: 1.96 A
Q14Mixed
A solid wire of radius 4 cm carries 20 A uniformly distributed. The field at a point 2 cm from the axis is:
(a) 2.5 × 10⁻⁵ T
(b) 5 × 10⁻⁵ T
(c) 1 × 10⁻⁴ T
(d) 2 × 10⁻⁴ T
Show step-by-step solution
Givena = 4 cm = 0.04 m, I = 20 A, r = 2 cm = 0.02 m
AskedField B
ConceptThe point is INSIDE the conductor, so use the inside formula, not the ordinary wire formula.4.6 Thick wire
FormulaB = (2 × 10⁻⁷) I r / a²
Solutionr = 0.02 m < a = 0.04 m ⇒ inside the wire. B = 2 × 10⁻⁷ × 20 × 0.02 / (0.04)² = 2 × 10⁻⁷ × 0.4 / 1.6 × 10⁻³ = 2 × 10⁻⁷ × 250 = 5 × 10⁻⁵ T
Answer: 5 × 10⁻⁵ T
Q15Mixed
A galvanometer of resistance 20 Ω gives full-scale deflection for 5 mA. The resistance needed in series to convert it into a voltmeter of range 10 V is:
(a) 980 Ω
(b) 1980 Ω
(c) 2000 Ω
(d) 2020 Ω
Show step-by-step solution
GivenG = 20 Ω, I_g = 5 mA = 0.005 A, V = 10 V
AskedSeries resistance R
ConceptThe range is given in volts ⇒ voltmeter conversion ⇒ high resistance in series.4.10 Galvanometer
A long straight wire lies outside a closed Amperian loop. For that loop:
(a) both ∮B·dl and B are zero
(b) ∮B·dl = 0 but B ≠ 0
(c) ∮B·dl ≠ 0 but B = 0
(d) both are non-zero
Show step-by-step solution
GivenCurrent-carrying wire outside the Amperian loop
AskedValues of the integral and of the field
ConceptAmpere's law counts only ENCLOSED current, but the field itself fills all of space.4.6 Ampere's law
Formula∮ B·dl = μ₀ I_enclosed
SolutionNo current is enclosed, so the integral is zero. But the wire still produces a field at every point on the loop. So the integral is zero while B is not.
Answer: ∮B·dl = 0 but B ≠ 0
Q18Mixed
A proton and an alpha particle with the same kinetic energy enter a magnetic field perpendicularly. The ratio of the radii of their paths is:
ConceptThe fixed quantity is kinetic energy, so use r = √(2mK)/qB — not the ordinary mv/qB.4.3 Circular motion
Formular = √(2 m K) / q B ⇒ r ∝ √m / q
SolutionProton: √m / q Alpha: √(4m) / 2q = 2√m / 2q = √m / q Identical, so the ratio is 1 : 1.
Answer: 1 : 1
Q19Mixed
A wire carrying current I is bent into one circular turn of radius R, giving field B at the centre. Re-bent into n turns, the field at the centre becomes:
(a) nB
(b) n²B
(c) B/n
(d) B/n²
Show step-by-step solution
GivenSame wire, 1 turn → n turns, same current
AskedNew field at the centre
ConceptThe wire length is fixed, so the radius shrinks to R/n. Track both N and R through the formula.4.5 Circular coil
FormulaB = μ₀ N I / 2R
Solution2πR = n × 2πr ⇒ r = R/n B′ = μ₀ n I / 2(R/n) = μ₀ n² I / 2R B′ = n² B
Answer: n²B
Q20Mixed
In a velocity selector the electric field is 6 × 10⁴ V/m and the magnetic field is 0.2 T. Particles pass undeflected if their speed is:
(a) 3 × 10⁴ m/s
(b) 3 × 10⁵ m/s
(c) 1.2 × 10⁴ m/s
(d) 1.2 × 10⁵ m/s
Show step-by-step solution
GivenE = 6 × 10⁴ V/m, B = 0.2 T
AskedSelected speed v
ConceptUndeflected means the electric and magnetic forces cancel exactly.4.2.2 Crossed fields
Formulaq E = q v B ⇒ v = E / B
Solutionv = 6 × 10⁴ / 0.2 = 3 × 10⁵ m/s
Answer: 3 × 10⁵ m/s
Q21Mixed
A solenoid 40 cm long with 200 turns carries 2 A. The field at its centre is:
(a) 6.28 × 10⁻⁴ T
(b) 1.256 × 10⁻³ T
(c) 2.51 × 10⁻³ T
(d) 5.02 × 10⁻⁴ T
Show step-by-step solution
GivenL = 40 cm = 0.4 m, N = 200, I = 2 A
AskedField at the centre
ConceptTotal turns and a length are given, so convert to turns per metre first — the commonest slip in solenoid questions.4.7 Solenoid
Formulan = N / L; B = μ₀ n I
Solutionn = 200 / 0.4 = 500 turns per metre B = 4π × 10⁻⁷ × 500 × 2 = 4π × 10⁻⁷ × 1000 = 1.256 × 10⁻³ T
Answer: 1.256 × 10⁻³ T
Q22Mixed
Two circular coils carry the same current and have radii in the ratio 2 : 3. The ratio of their magnetic moments is:
(a) 2 : 3
(b) 4 : 9
(c) 3 : 2
(d) 9 : 4
Show step-by-step solution
GivenSame current, r₁ : r₂ = 2 : 3
AskedRatio of magnetic moments
ConceptWith current fixed, the moment depends only on area, which goes as the square of the radius.4.9 Torque
Formulam = I πr² ⇒ m ∝ r²
Solutionm₁ / m₂ = r₁² / r₂² = 4 / 9 Ratio = 4 : 9
Answer: 4 : 9
Q23Mixed
The net force on a closed current-carrying loop placed in a uniform magnetic field is:
(a) BIL
(b) zero
(c) IAB
(d) 2BIL
Show step-by-step solution
GivenClosed loop in a uniform field
AskedNet force
ConceptA closed loop has zero effective length, since it begins and ends at the same point.4.2.3 Force on a wire
FormulaF = I L_eff B, with L_eff = 0
SolutionThe straight-line distance between the start and end of a closed loop is zero. So the net force is zero. A torque may still act — that is a different quantity.
Answer: zero
Q24Mixed
The magnetic field due to a long straight wire at distance r is B. At distance 3r it becomes:
(a) B/3
(b) B/9
(c) 3B
(d) 9B
Show step-by-step solution
GivenField B at distance r
AskedField at distance 3r
ConceptA straight wire's field falls as 1/r, not 1/r² — the trap in this question.4.6 Straight wire
FormulaB ∝ 1 / r
SolutionB′ / B = r / 3r = 1/3 B′ = B / 3
Answer: B/3
Q25Mixed
In a moving coil galvanometer, the field is made radial so that:
(a) the coil rotates faster
(b) the scale is linear
(c) the resistance is reduced
(d) the current sensitivity is halved
Show step-by-step solution
GivenRadial field produced by curved poles and a soft iron core
AskedPurpose of the radial field
ConceptIn a radial field the coil plane always stays along B, so sin θ = 1 at every position.4.10 Galvanometer
Formulaτ = N I A B sin θ, with sin θ = 1
SolutionWithout a radial field, τ would vary as sin θ as the coil turned. A radial field keeps sin θ = 1 always, so τ = NIAB and θ ∝ I. Equal current steps then give equal deflection steps — a linear scale.
Answer: the scale is linear
Q26Mixed
A charged particle enters a uniform magnetic field at 45° to it. Its path is:
(a) a straight line
(b) a circle
(c) a helix
(d) a parabola
Show step-by-step solution
GivenEntry angle 45°
AskedShape of the path
ConceptSplit the velocity into an across-field part (which circles) and an along-field part (which drifts).4.3 Helix
Formulav⊥ = v sin θ, v∥ = v cos θ
Solutionv sin 45° across the field bends into a circle. v cos 45° along the field carries it forward steadily. Circle plus steady drift = a helix.
Answer: a helix
Q27Mixed
The magnetic field due to a current element at a point lying along the direction of the element is:
(a) maximum
(b) zero
(c) half the maximum
(d) infinite
Show step-by-step solution
GivenPoint along the element's own axis
AskedValue of dB
ConceptThe sin θ factor vanishes along the element's own line — the 'blind spot' of Biot–Savart.4.4 Biot–Savart
FormuladB = (μ₀/4π) I dl sin θ / r²
SolutionAlong the element's own direction, θ = 0°. sin 0° = 0 So dB = 0 there.
Answer: zero
Q28Mixed
A circular loop is fed at two diametrically opposite points so that the current divides equally between the two halves. The field at the centre is:
(a) μ₀I/2R
(b) μ₀I/4R
(c) μ₀I/8R
(d) zero
Show step-by-step solution
GivenCurrent divides equally into two semicircles
AskedField at the centre
ConceptEach half gives the same field magnitude, but the two currents circulate in opposite senses.4.5 Split loop
FormulaB_semicircle = μ₀I′ / 4R, added as vectors
SolutionEach half carries I/2, giving μ₀I/8R. One half goes clockwise, the other anticlockwise ⇒ their fields oppose. Equal and opposite ⇒ they cancel exactly: B = 0.
Answer: zero
Q29Mixed
A magnetic field can change which of the following for a charged particle?
(a) its speed
(b) its kinetic energy
(c) the direction of its velocity
(d) the magnitude of its momentum
Show step-by-step solution
GivenCharged particle in a magnetic field
AskedThe quantity that CAN change
ConceptThe magnetic force is always perpendicular to the velocity, so it does no work — it can only steer.4.2.2 Lorentz force
FormulaW = 0
SolutionNo work is done, so speed, kinetic energy and the magnitude of momentum are all unchanged. But the force does turn the particle. So only the DIRECTION of the velocity changes.
Answer: the direction of its velocity
Q30Mixed
One ampere is defined using two parallel wires 1 m apart carrying 1 A each, which experience a force per unit length of:
(a) 10⁻⁷ N/m
(b) 2 × 10⁻⁷ N/m
(c) 4π × 10⁻⁷ N/m
(d) 10⁻⁴ N/m
Show step-by-step solution
GivenI₁ = I₂ = 1 A, d = 1 m, vacuum
AskedForce per unit length
ConceptSubstitute the defining values into the two-wire formula, using μ₀/2π = 2 × 10⁻⁷.4.8 Ampere definition
Formulaf = μ₀ I₁ I₂ / 2πd
Solutionf = 2 × 10⁻⁷ × 1 × 1 / 1 = 2 × 10⁻⁷ N/m
Answer: 2 × 10⁻⁷ N/m
Q31Mixed
A current loop is in stable equilibrium in a uniform field when its magnetic moment is:
(a) parallel to B
(b) antiparallel to B
(c) perpendicular to B
(d) at 45° to B
Show step-by-step solution
GivenLoop free to rotate in a uniform field
AskedCondition for stable equilibrium
ConceptBoth 0° and 180° give zero torque, but only one restores the loop after a nudge.4.9 Torque
Formulaτ = mB sin θ; U = − mB cos θ
SolutionAt θ = 0°: U = −mB, the minimum energy — a nudge brings it back ⇒ stable. At θ = 180°: U = +mB, the maximum — a nudge flips it over ⇒ unstable. So stable equilibrium is with m parallel to B.
Answer: parallel to B
Q32Mixed
For a solid cylindrical conductor of radius a carrying a uniform current, the graph of B against distance r from the axis:
(a) decreases as 1/r throughout
(b) rises linearly up to r = a, then falls as 1/r
(c) is constant inside and falls as 1/r outside
(d) rises as r² up to r = a, then falls as 1/r²
Show step-by-step solution
GivenSolid conductor of radius a, uniform current
AskedShape of the B versus r graph
ConceptInside, only part of the current is enclosed; outside, all of it is.4.6 Thick wire
FormulaInside B ∝ r; Outside B ∝ 1/r
SolutionInside: B = μ₀Ir/2πa², a straight line from zero. At r = a the field is maximum. Outside: B = μ₀I/2πr, a 1/r decay. So: linear rise, then 1/r fall.
Answer: rises linearly up to r = a, then falls as 1/r
Q33Mixed
An electron with specific charge e/m = 1.76 × 10¹¹ C/kg moves in a field of 3.57 × 10⁻² T. Its frequency of revolution is:
(a) 1 MHz
(b) 100 MHz
(c) 1 GHz
(d) 10 GHz
Show step-by-step solution
Givene/m = 1.76 × 10¹¹ C/kg, B = 3.57 × 10⁻² T
AskedFrequency ν
ConceptAnswer in hertz for a charge in a field ⇒ cyclotron frequency, which is independent of speed.4.3 Circular motion
The magnetic field at one end of a long solenoid, compared with the field at its centre, is:
(a) the same
(b) half
(c) double
(d) zero
Show step-by-step solution
GivenPoint on the axis at the end of a long solenoid
AskedField at the end
ConceptAt the end, only half the winding lies to one side, so the 'very long both ways' assumption fails.4.7 Solenoid
FormulaB_end = μ₀ n I / 2
SolutionAt the centre B = μ₀nI. At the end only one side contributes ⇒ B = μ₀nI/2. So it is exactly half.
Answer: half
Q35Mixed
In the formula F = I l × B, the field B refers to:
(a) the field produced by the wire itself
(b) an external magnetic field
(c) the sum of both fields
(d) the Earth's field only
Show step-by-step solution
GivenForce on a current-carrying conductor
AskedMeaning of B
ConceptA wire cannot push itself, so the field must come from outside.4.2.3 Force on a wire
FormulaF = I l × B
SolutionThe wire's own circular field cannot exert a net force on the wire. Just as you cannot lift yourself by your own shoelaces. So B must be an external field.
Answer: an external magnetic field
Q36Mixed
Far along the axis of a small current loop (x ≫ R), the magnetic field varies as:
(a) 1/x
(b) 1/x²
(c) 1/x³
(d) 1/x⁴
Show step-by-step solution
GivenAxial point with x ≫ R
AskedDependence on x
ConceptDrop R² compared with x² in the axial formula — the dipole signature emerges.4.5 Coil axis
FormulaB = μ₀IR² / 2(x²+R²)^{3/2}
SolutionFor x ≫ R, (x² + R²)^{3/2} ≈ x³ B ≈ μ₀IR²/2x³ So B ∝ 1/x³.
Answer: 1/x³
Q37Mixed
Doubling the number of turns in a galvanometer coil:
(a) doubles both sensitivities
(b) doubles the current sensitivity but may leave the voltage sensitivity unchanged
(c) doubles the voltage sensitivity only
(d) changes neither
Show step-by-step solution
GivenNumber of turns doubled
AskedEffect on the two sensitivities
ConceptMore turns also means more wire and therefore more coil resistance G, which appears only in the voltage sensitivity.4.10 Galvanometer
FormulaI_s = NAB/k; V_s = NAB/kG
SolutionCurrent sensitivity ∝ N, so it doubles. Voltage sensitivity = NAB/kG, and G roughly doubles as well. The two effects cancel, so the voltage sensitivity may not increase.
Answer: doubles the current sensitivity but may leave the voltage sensitivity unchanged
Q38Mixed
A coil of 50 turns and area 4 × 10⁻² m² carries 2 A in a field of 0.5 T with its plane perpendicular to the field. The torque on it is:
(a) 0
(b) 1 N m
(c) 2 N m
(d) 4 N m
Show step-by-step solution
GivenN = 50, A = 4 × 10⁻² m², I = 2 A, B = 0.5 T, plane perpendicular to B
AskedTorque τ
ConceptPlane perpendicular to B means the NORMAL is along B, so θ = 0° — the trap in this question.4.9 Torque
Formulaτ = N I A B sin θ
SolutionPlane ⊥ B ⇒ normal ∥ B ⇒ θ = 0° sin 0° = 0 τ = 0
Answer: 0
Q39Mixed
The force per unit length between two parallel wires varies with their separation d as:
(a) 1/d
(b) 1/d²
(c) d
(d) d²
Show step-by-step solution
GivenTwo long parallel current-carrying wires
AskedDependence on separation
ConceptRead it from the formula — d appears once in the denominator.4.8 Parallel wires
Formulaf = μ₀I₁I₂ / 2πd
Solutionf ∝ 1/d. This is NOT an inverse-square law, unlike Coulomb's law between charges.
Answer: 1/d
Q40Mixed
The work done by the magnetic force on a charged particle completing one full circle is:
(a) 2πrqvB
(b) qvB
(c) zero
(d) ½mv²
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GivenOne complete revolution in a magnetic field
AskedWork done
ConceptThe force is perpendicular to the displacement at every instant.4.3 Circular motion
FormulaW = F d cos 90° = 0
SolutionAt every point the magnetic force is perpendicular to the velocity. Each small element of work is zero. Total work over the revolution = zero.
Answer: zero
Q41Mixed
A long straight wire is bent into a semicircle of radius R at its middle, with the two straight portions lying along the diameter. The field at the centre is:
(a) μ₀I/2R
(b) μ₀I/4R
(c) μ₀I/8R
(d) μ₀I/2πR
Show step-by-step solution
GivenSemicircular arc of radius R with straight portions along the diameter
AskedField at the centre
ConceptThe straight parts lie along the line to the centre, so they contribute nothing.4.5 Semicircle
FormulaB = μ₀I / 4R
SolutionFor the straight sections, dl and r are parallel ⇒ sin θ = 0 ⇒ no contribution. Only the semicircle counts, giving half a full loop. B = μ₀I/4R
Answer: μ₀I/4R
Q42Mixed
A galvanometer of resistance 40 Ω is shunted by 10 Ω. The fraction of the total current passing through the galvanometer is:
(a) 1/5
(b) 1/4
(c) 1/2
(d) 4/5
Show step-by-step solution
GivenG = 40 Ω, S = 10 Ω
AskedFraction I_g / I
ConceptCurrent divides between parallel paths in inverse proportion to their resistances.4.10 Galvanometer
FormulaI_g / I = S / (S + G)
SolutionI_g / I = 10 / (10 + 40) = 10 / 50 = 1/5
Answer: 1/5
Q43Mixed
An electron and a proton enter the same magnetic field with the same velocity, perpendicular to the field. They will:
(a) bend the same way
(b) bend in opposite directions
(c) travel in straight lines
(d) come to rest
Show step-by-step solution
GivenElectron and proton, same v, same B, perpendicular entry
AskedDirections of bending
ConceptThe sign of the charge decides the sign of the force.4.2.2 Lorentz force
FormulaF = q (v × B)
SolutionFor the proton (positive), F is along v × B. For the electron (negative), F is opposite to v × B. So they bend in opposite directions.
Answer: bend in opposite directions
Q44Mixed
The relation between ε₀, μ₀ and the speed of light c is:
(a) ε₀μ₀ = c
(b) ε₀μ₀ = c²
(c) ε₀μ₀ = 1/c
(d) ε₀μ₀ = 1/c²
Show step-by-step solution
GivenPermittivity, permeability and the speed of light
AskedThe correct relation
ConceptThe electric and magnetic constants together produce the speed of light — Maxwell's clue that light is electromagnetic.4.4 Constants
Which of the following does NOT depend on the radius of the current configuration?
(a) field at the centre of a circular coil
(b) field inside a long solenoid
(c) magnetic moment of a coil
(d) field on the axis of a coil
Show step-by-step solution
GivenFour standard configurations
AskedThe one independent of radius
ConceptCheck each formula and see where R appears.4.5 / 4.7 Recognition
FormulaB = μ₀NI/2R; B = μ₀nI; m = NIπR²
SolutionCoil centre: B = μ₀NI/2R — contains R. Solenoid: B = μ₀nI — no R anywhere. Magnetic moment: m = NIπR² — contains R. Coil axis: contains R too. So the solenoid field is the one independent of radius.
Answer: field inside a long solenoid
Q46Mixed
Assertion (A): Two parallel wires carrying currents in the same direction attract each other. Reason (R): Like electric charges also attract each other.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenParallel currents and like charges
AskedTruth of A and R
ConceptThe first statement is a genuine result of this chapter; the second is a false analogy.Assertion–Reason
Formulaf = μ₀I₁I₂/2πd; Coulomb's law
SolutionA is TRUE: parallel currents do attract. R is FALSE: like charges REPEL, they do not attract. So A is true but R is false. (This reversal is exactly what makes the topic tricky.)
Answer: A is true but R is false
Q47Mixed
Assertion (A): The kinetic energy of a charged particle moving in a uniform magnetic field remains constant. Reason (R): The magnetic force is always perpendicular to the velocity.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Show step-by-step solution
GivenCharged particle in a uniform magnetic field
AskedTruth of A and R, and whether R explains A
ConceptBoth are true, and the second is precisely the reason for the first.Assertion–Reason
FormulaW = 0 ⇒ KE constant
SolutionA is TRUE: the kinetic energy never changes. R is TRUE: the force is always perpendicular to v. Perpendicular force ⇒ no work done ⇒ energy unchanged. So R explains A.
Answer: Both A and R are true, and R is the correct explanation of A
Q48Mixed
Match each configuration with its field:
(a) Long straight wire (b) Inside a long solenoid (c) Centre of a semicircular arc (d) Inside a hollow pipe
(i) μ₀nI (ii) zero (iii) μ₀I/2πr (iv) μ₀I/4R
(a) a–iii, b–i, c–iv, d–ii
(b) a–iii, b–iv, c–i, d–ii
(c) a–i, b–iii, c–iv, d–ii
(d) a–iii, b–i, c–ii, d–iv
Show step-by-step solution
GivenFour standard configurations
AskedCorrect matching
ConceptThis is the recognition drill in question form — the skill the whole test is designed to measure.Match the following
FormulaStandard field formulas
Solution(a) Long straight wire ⇒ μ₀I/2πr ⇒ (iii) (b) Inside a solenoid ⇒ μ₀nI ⇒ (i) (c) Semicircular arc ⇒ μ₀I/4R ⇒ (iv) (d) Inside a hollow pipe ⇒ zero ⇒ (ii)
Answer: a–iii, b–i, c–iv, d–ii
Q49Mixed
Match each quantity with its SI unit:
(a) Magnetic moment (b) Magnetic field (c) Force between two wires (d) Torque on a loop
(i) tesla (ii) N m (iii) A m² (iv) N per metre
(a) a–iii, b–i, c–iv, d–ii
(b) a–i, b–iii, c–iv, d–ii
(c) a–iii, b–i, c–ii, d–iv
(d) a–ii, b–i, c–iv, d–iii
Show step-by-step solution
GivenFour quantities from across the chapter
AskedCorrect matching
ConceptIdentifying a formula by the unit of its answer is the fastest triage in the exam.Match the following
Formulam = IA; B in tesla; f in N/m; τ in N m
Solution(a) Magnetic moment ⇒ A m² ⇒ (iii) (b) Magnetic field ⇒ tesla ⇒ (i) (c) Force per unit length between wires ⇒ N per metre ⇒ (iv) (d) Torque ⇒ N m ⇒ (ii)
Answer: a–iii, b–i, c–iv, d–ii
Q50Mixed
Which of the following statements is INCORRECT?
(a) A magnetic field does no work on a moving charge
(b) The time period of circular motion in a magnetic field is independent of speed
(c) The field inside a long solenoid depends on its radius
(d) The net force on a closed loop in a uniform field is zero
Show step-by-step solution
GivenStatements from across the chapter
AskedThe incorrect statement
ConceptThree of these are core results; one contradicts a formula that contains no radius term.Recognition · summary
FormulaW = 0; T = 2πm/qB; B = μ₀nI; F_net = 0
Solution'No work' — correct, the force is always perpendicular to v. 'T independent of speed' — correct, T = 2πm/qB. 'Solenoid field depends on radius' — INCORRECT. B = μ₀nI contains no radius. 'Zero net force on a closed loop in a uniform field' — correct.
Answer: The field inside a long solenoid depends on its radius
After the test — how to read the result
Every solution above carries a small grey tag naming the section it came from. Sort the wrong answers by that tag and the diagnosis writes itself.
Section tags on the errors
What it means
What to do
Two or more from 4.9
Torque and moment are weak
Redo Priority 1 pack
Two or more from 4.4–4.7
Field formulas are weak
Redo Priority 2 pack
Two or more from 4.2.2 or 4.3
Charged particle motion is weak
Redo Priority 3 pack
Two or more from 4.2.3 or 4.8
Forces on conductors are weak
Redo Priority 4 pack
Two or more from 4.10
Galvanometer is weak
Redo Priority 5 pack
Scattered singles across many sections
Concepts are fine; the losses are arithmetic and unit slips
Drill conversions and timing, not theory
Right answers but slow, or many guesses
Recognition is the bottleneck
Redo Part 1 keyword decoder, then retake this test
The recognition drill, without any calculation. Go back through all 50 questions and, for each one, say only which formula it needs — no numbers, no working. Two minutes for the lot. If she names 45 or more correctly, the chapter is secure and the remaining marks are speed. If she names fewer than 40, that is the real problem, and no amount of extra numerical practice will fix it.