Moving Charges and Magnetism NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Moving Charges and Magnetism question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Moving Charges and Magnetism previous year questions
36 questions. Tap an option to check your answer.
Q1NEET 2021
In the product F⃗ = q(v⃗ × B⃗) = q v⃗ × (B î + B ĵ + B₀ k̂) For q = 1 and v⃗ = 2î + 4ĵ + 6k̂ and F⃗ = 4î − 20ĵ + 12k̂ What will be the complete expression for B⃗ ?
Not quite — the answer is B.
With B = Bî + Bĵ + B₀k̂ and v = 2î + 4ĵ + 6k̂: v × B = (4B₀ − 6B)î + (6B − 2B₀)ĵ + (2B − 4B)k̂. Matching the k̂ term, −2B = 12 so B = −6; matching î, 4B₀ + 36 = 4 so B₀ = −8. Check ĵ: −36 + 16 = −20, so B = −6î − 6ĵ − 8k̂.
Q2NEET 2021
A uniform conducting wire of length 12a and resistance 'R' is wound up as a current carrying coil in the shape of, (i) an equilateral triangle of side 'a'. (ii) a square of side 'a'. The magnetic dipole moments of the coil in each case respectively are :
Not quite — the answer is A.
Triangle: each turn uses 3a, so 12a gives 4 turns of area (√3/4)a², and m = 4I(√3/4)a² = √3 Ia². Square: each turn uses 4a, so 3 turns of area a², and m = 3Ia².
Q3NEET 2020 · Q172
A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : (µ₀ = 4π×10⁻⁷ T m A⁻¹)
Not quite — the answer is A.
Turns per metre n = 100/0.5 = 200, so B = µ₀nI = 4π×10⁻⁷ × 200 × 2.5 = 2π×10⁻⁴ ≈ 6.28×10⁻⁴ T. Using the total turns per 1 m (100) instead of per 0.5 m halves the answer to 3.14×10⁻⁴ T.
Q4NEET 2019 · Q3
Ionized hydrogen atoms and α-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r_H : r_α will be:
Not quite — the answer is A.
r = p/(qB); with equal momenta r ∝ 1/q. The proton has charge e and the α-particle 2e, so r_H : r_α = 2 : 1. Taking 1 : 2 inverts the dependence on charge.
Q5NEET 2018 · Q26
A metallic rod of mass per unit length 0.5 kg m⁻¹ is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
Not quite — the answer is D.
The magnetic force BIl is horizontal; balancing components along the incline: BIl cos 30° = mg sin 30°. I = (m/l)g tan 30°/B = 0.5 × 9.8 × 0.577/0.25 ≈ 11.32 A. Taking the force along the incline (without the cos 30°) would give a smaller current.
Q6NEET 2018 · Q29
Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
Not quite — the answer is A.
Voltage sensitivity = current sensitivity / G, so G = (5 div/mA)/(20 div/V) = (5000 div/A)/(20 div/V) = 250 Ω. Forgetting the mA-to-A conversion gives 0.25 Ω-type answers.
Q7NEET 2016 Phase 2 · Q27
A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be
Not quite — the answer is B.
With the same wire, n turns means each turn has radius r/n. B′ = μ₀nI/(2(r/n)) = n² × μ₀I/(2r) = n²B. nB forgets that the radius also shrinks by a factor of n.
Q8NEET 2016 Phase 2 · Q29
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10⁻² T. If the value of e/m is 1.76 × 10¹¹ C/kg, the frequency of revolution of the electron is
Not quite — the answer is A.
f = (e/m)B/(2π) = (1.76 × 10¹¹ × 3.57 × 10⁻²)/(2π) = 6.28 × 10⁹/6.28 ≈ 1 × 10⁹ Hz = 1 GHz. 6.28 × 10⁹ is the angular frequency ω, so dividing by 2π is essential; 100 MHz is off by a factor of 10.
Q9NEET 2016 Phase 1 · Q26
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distances a/2 and 2a respectively, from the axis of the wire is:
Not quite — the answer is C.
Inside the wire B = μ₀Ir/(2πa²), so at a/2, B = μ₀I/(4πa). Outside B' = μ₀I/(2πr), so at 2a, B' = μ₀I/(4πa). The ratio is 1. Treating the inside point as if all the current were enclosed (using 1/r there) would wrongly give a ratio of 4.
Q10AIPMT 2015
An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced at the centre has magnitude:
Not quite — the answer is A.
The orbiting electron is a current loop with I = charge passing per second = ne. The field at the centre of a loop is μ₀I/(2r) = μ₀ne/(2r). μ₀ne/(2πr) is the formula for a long straight wire.
Q11AIPMT 2014
Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that 'O' is their common point for the two. The wires carry I₁ and I₂ currents respectively. Point 'P' is lying at distance 'd' from 'O' along a direction perpendicular to the plane containing the wires. The magnetic field at the point 'P' will be:
Not quite — the answer is C.
P is at distance d from each long wire, so each gives B = μ₀I/(2πd). The two wires are perpendicular, so their fields at P are perpendicular and add as vectors: B = (μ₀/2πd)√(I₁² + I₂²), not the simple sum.
Q12AIPMT 2014
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be:
Not quite — the answer is B.
The galvanometer carries I/500, and since it is in parallel with the shunt, the voltage across the ammeter is (I/500)G. Ammeter resistance = V/I = G/500; G/499 is the shunt alone, not the combination.
Q13NEET 2013
When a proton is released from rest in a room, it starts with an initial acceleration a₀ towards west. When it is projected towards north with a speed v₀ it moves with an initial acceleration 3a₀ towards west. The electric and magnetic fields in the room are
Not quite — the answer is B.
At rest only the electric force acts: eE = ma₀ westward, so E = ma₀/e towards west. Moving north, the extra 2ma₀ westward comes from the magnetic force: ev₀B = 2ma₀, B = 2ma₀/(ev₀). For a positive charge, v (north) × B must point west, which happens when B points down.
Q14NEET 2013
A current loop in a magnetic field -
Not quite — the answer is D.
Torque on a loop is τ = m × B, which is zero when the magnetic moment is parallel or antiparallel to B. The parallel position (minimum energy) is stable and the antiparallel position (maximum energy) is unstable. A loop does not feel a torque in every orientation.
Q15AIPMT 2012 Mains
A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α-particle to describe a circle of same radius in the same field?
Not quite — the answer is B.
r = √(2mK)/(qB), so for the same r and B, K ∝ q²/m. For an α-particle q²/m = (2e)²/(4m_p) = e²/m_p, the same as the proton, so it needs the same 1 MeV.
Q16AIPMT 2012 Prelims
Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be
Not quite — the answer is B.
The coils give fields μ₀I/(2R) and μ₀(2I)/(2R) = μ₀I/R at the centre, and these are perpendicular because the coil planes are perpendicular. Resultant B = √[(μ₀I/2R)² + (μ₀I/R)²] = √5 μ₀I/(2R). Simply adding them (3μ₀I/2R) ignores that the fields are at right angles.
Q17AIPMT 2012 Prelims
A millivoltmeter of 25 millivolt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be
Not quite — the answer is B.
The meter gives full deflection at 25 mV across it; with the shunt carrying almost all 25 A, the shunt resistance is S ≈ V/I = 25 × 10⁻³/25 = 10⁻³ Ω = 0.001 Ω. 0.01 Ω would pass only 2.5 A at 25 mV.
Q18AIPMT 2011 Mains
Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency f Hz. The magnitude of magnetic induction at the center of the ring is
Not quite — the answer is C.
The rotating ring is equivalent to a current I = q/T = qf. The field at the centre of a circular loop is μ₀I/(2R) = μ₀qf/(2R). The expression with 2πR in the denominator is the formula for a long straight wire, not a loop.
Q19AIPMT 2011 Prelims
A current carrying closed loop in the form of a right angle isosceles triangle ABC (right angle at B) is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F⃗, the force on the arm AC is
Not quite — the answer is C.
The field is along AB, so the force on arm AB is zero. The net magnetic force on any closed current loop in a uniform field is zero, so F(BC) + F(AC) = 0 and the force on AC is −F⃗. A factor of √2 is the trap from the longer length of AC, but only the component of AC perpendicular to B (equal to BC) matters.
Q20AIPMT 2011 Prelims
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron
Not quite — the answer is C.
Since the velocity is parallel to B, the magnetic force qv × B is zero and there is no sideways turning. The electron's charge is negative, so the electric force is opposite to E, i.e. opposite to its velocity, and it decelerates, so its speed decreases.
Q21AIPMT 2010 Mains · Q61
A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts at their common centre is
Not quite — the answer is A.
Each semicircle gives μ₀i/(4R) at the centre, half the field of a full loop. Because the two semicircles lie in perpendicular planes, their fields are perpendicular, so the resultant is √2 × μ₀i/(4R) = μ₀i/(2√2 R). Simply adding the two fields gives μ₀i/(2R), which wrongly treats them as parallel.
Q22AIPMT 2010 Prelims · Q29
A galvanometer has a coil of resistance 100 ohm and gives a full scale deflection for 30 mA current. If it is to work as a voltmeter of 30 volt range, the resistance required to be added will be
Not quite — the answer is B.
Total resistance needed = V/I_g = 30/0.030 = 1000 Ω. Series resistance = 1000 − 100 = 900 Ω. 1000 Ω forgets to subtract the coil's own resistance.
Q23AIPMT 2010 Prelims · Q32
A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is F⃗, the net force on the remaining three arms of the loop is
Not quite — the answer is C.
The net magnetic force on a closed current loop in a uniform field is zero. So the other three arms together must give −F⃗, cancelling the force on the first arm.
Q24AIPMT 2010 Prelims · Q33
A thin ring of radius R meter has charge q coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of f revolutions/s. The value of magnetic induction in Wb/m² at the centre of the ring is
Not quite — the answer is A.
The rotating charge acts as a current I = qf. Field at the centre of a ring is B = μ₀I/(2R) = μ₀qf/(2R). Putting 2πR in the denominator confuses it with the field of a long straight wire.
Q25AIPMT 2010 Prelims · Q40
A beam of cathode rays is subjected to crossed electric (E) and magnetic (B) fields. The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by (where V is the potential difference between cathode and anode)
Not quite — the answer is A.
No deflection means eE = evB, so v = E/B. Accelerated through V: eV = ½mv², so e/m = v²/(2V) = E²/(2VB²). The inverted forms come from writing v = B/E.
Q26AIPMT 2009 Prelims · Q24
A galvanometer having a coil resistance of 60 Ω shows full scale deflection when a current of 1.0 amp passes through it. It can be converted into an ammeter to read currents upto 5.0 amp by:
Not quite — the answer is C.
Shunt S = IgG/(I − Ig) = (1 × 60)/(5 − 1) = 15 Ω connected in parallel, so the extra 4 A bypasses the coil. A series resistance would make a voltmeter, not an ammeter.
Q27AIPMT 2009 Prelims · Q43
The magnetic force acting on a charged particle of charge −2 μC in a magnetic field of 2 T acting in y direction, when the particle velocity is (2î + 3ĵ) × 10⁶ m s⁻¹, is:
Not quite — the answer is D.
F⃗ = q(v⃗ × B⃗). Only the x component of velocity matters: (2 × 10⁶ î) × (2ĵ) = 4 × 10⁶ k̂. Multiplying by q = −2 × 10⁻⁶ C gives −8k̂ N, i.e. 8 N in −z direction. Ignoring the negative charge gives +z.
Q28AIPMT 2009 Prelims · Q48
Under the influence of a uniform magnetic field, a charged particle moves with constant speed V in a circle of radius R. The time period of rotation of the particle:
Not quite — the answer is B.
qVB = mV²/R gives R = mV/qB, and T = 2πR/V = 2πm/qB. Since R grows in proportion to V, T depends on neither speed nor radius.
Q29AIPMT 2008 Prelims · Q33
A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of induction B⃗. After 3 seconds the kinetic energy of the particle will be
Not quite — the answer is D.
The magnetic force qv × B is always perpendicular to the velocity, so it does no work. The particle moves in a circle at constant speed and its kinetic energy stays T. The 3 seconds is a distractor: the elapsed time does not matter.
Q30AIPMT 2007 Prelims · Q2
A beam of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move
Not quite — the answer is A.
With only the magnetic field, the force qv × B is always perpendicular to the velocity, which is itself perpendicular to B, so the electrons move in a circle. A parabolic path needs a constant force such as an electric field, which has been switched off.
Q31AIPMT 2007 Prelims · Q10
In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular path of radius R using a magnetic field B. If V and B are kept constant, the ratio (charge on the ion/mass of the ion) will be proportional to
Not quite — the answer is A.
qV = ½mv² and R = mv/(qB) give R² = 2mV/(qB²), so q/m = 2V/(B²R²) ∝ 1/R². Answering 1/R ignores that v itself depends on q/m.
Q32AIPMT 2007 Prelims · Q34
A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment μ is given by
Not quite — the answer is D.
The circulating charge is a current I = q/T = qv/(2πR); μ = I × πR² = qvR/2. qvR misses the factor ½.
Q33AIPMT 2007 Prelims · Q47
Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius R with constant speed v. The time period of the motion
Not quite — the answer is B.
T = 2πR/v = 2πm/(qB); a larger speed gives a proportionally larger radius, so the period depends on neither R nor v.
Q34AIPMT 2006 Prelims · Q5
When a charged particle moving with velocity v⃗ is subjected to a magnetic field of induction B⃗, the force on it is non-zero. This implies that :
Not quite — the answer is C.
The magnetic force is F = qvB sin θ, which vanishes only when θ = 0° or 180°. So a non-zero force tells us only that θ is anything other than 0° and 180°. It need not be exactly 90°; 90° simply gives the maximum force.
Q35AIPMT 2005 Prelims · Q1
A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet such that B⃗ is in plane of the coil. If due to a current i in the triangle a torque τ acts on it, the side l of the triangle is :
Not quite — the answer is C.
With B in the plane of the coil the torque is maximum: τ = iAB with A = (√3/4)l². So l² = 4τ/(√3 Bi) and l = 2(τ/(√3 Bi))^(1/2). The trap (2/√3)(τ/Bi)^(1/2) wrongly keeps √3 outside the square root.
Q36AIPMT 2005 Prelims · Q43
An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the circle. The radius of the circle is proportional to :
Not quite — the answer is C.
The orbiting electron is a current I = ev/(2πr), giving B = μ₀I/(2r) = μ₀ev/(4πr²). So r² ∝ v/B and r ∝ √(v/B). v/B would follow from B ∝ v/r, which ignores that the current itself depends on r.
How often Moving Charges and Magnetism came up
Questions from this chapter in each paper on this site.
- NEET 2021: 2
- NEET 2020: 1
- NEET 2019: 1
- NEET 2018: 2
- NEET 2016 Phase 2: 2
- NEET 2016 Phase 1: 1
- AIPMT 2015: 1
- AIPMT 2014: 2
- NEET 2013: 2
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 2
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 2
- AIPMT 2010 Mains: 1
- AIPMT 2010 Prelims: 4
- AIPMT 2009 Prelims: 3
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 4
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 2
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