68 free MCQs on Moving Charges and Magnetism with worked answers and explanations. Magnetic force on moving charges and currents, the Biot-Savart law, Ampere's law, and the cyclotron - how electric currents create and respond to magnetic fields.
Take the timed Moving Charges and Magnetism chapterwise test →Below are 68 practice questions on Moving Charges and Magnetism, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Moving Charges and Magnetism notes.
The magnetic field around a long straight current-carrying wire forms concentric circles, with direction given by the right-hand rule (point thumb along I, fingers curl in the direction of B).
Q1.
The SI unit of magnetic field B is:
Answer: C. Tesla
Why: Magnetic field strength is measured in Tesla (T). 1 T = 1 N/(A m).
Q2.
Force on a charge q moving with velocity v in magnetic field B:
Answer: C. F = qvB sin(theta)
Why: F = qvB sin(theta) where theta is angle between v and B. Maximum when v is perpendicular to B (sin90=1).
Q3.
A current-carrying conductor placed in a magnetic field experiences a force. This is called:
Answer: B. Ampere force
Why: Ampere (or motor) force: F = BIL sin(theta). A current-carrying conductor in a magnetic field experiences this force.
Q4.
Direction of magnetic field inside a solenoid:
Answer: A. From south to north inside
Why: Inside a solenoid, field lines run from south to north (from one end to the other, along the axis).
Q5.
A moving coil galvanometer uses a:
Answer: C. Uniform radial magnetic field
Why: A moving coil galvanometer uses a uniform radial magnetic field so that the torque is directly proportional to current at all deflections.
Q6.
In an electric motor, electrical energy converts to:
Answer: B. Mechanical energy
Why: An electric motor converts electrical energy into mechanical (kinetic) energy using the interaction of current and magnetic field.
Q7.
Oersted's experiment showed that:
Answer: B. Current produces magnetic field around a conductor
Why: Oersted showed in 1820 that an electric current creates a magnetic field around the conductor: linking electricity and magnetism.
Q8.
A charge moving parallel to a magnetic field:
Answer: B. Experiences minimum force of zero
Why: F = qvB sin(theta). When v is parallel to B, theta=0, sin(0)=0. Force = 0.
Q9.
A long, straight current-carrying wire produces a magnetic field whose field lines form:
Answer: A. Concentric circles around the wire
Why: The magnetic field around a long straight current-carrying wire forms concentric circles centred on the wire, with direction given by the right-hand rule.
Q10.
Which device uses the principle that a current-carrying coil placed in a magnetic field experiences a torque, to measure small electric currents?
Answer: B. Galvanometer
Why: A galvanometer is built on the principle that a current-carrying coil placed in a magnetic field experiences a torque proportional to the current, causing it to deflect and indicate the current's magnitude.
Q11.
A moving electric charge produces:
Answer: A. a magnetic field
Why: A charge in motion (a current) sets up a magnetic field around it.
Q12.
The SI unit of magnetic field (magnetic flux density) is the:
Answer: A. tesla
Why: Magnetic field B is measured in tesla (T).
Q13.
The magnetic force on a charge moving parallel to a magnetic field is:
Answer: A. zero
Why: F = qvB sinθ, and sin 0° = 0, so a charge moving along B feels no force.
Q14.
The direction of the force on a current-carrying conductor in a magnetic field is given by:
Answer: A. Fleming's left-hand rule
Why: Fleming's left-hand rule gives the direction of the motor force.
Q15.
A current-carrying conductor placed in a magnetic field experiences a:
Answer: A. force
Why: The interaction of the current with the field produces a force on the conductor.
Q16.
The magnetic field lines around a straight current-carrying wire form:
Answer: A. concentric circles
Why: The field lines are concentric circles centred on the wire.
Q17.
The instrument used to detect and measure small electric currents is the:
Answer: A. galvanometer
Why: A galvanometer detects and measures very small currents.
Q18.
The magnetic force on a stationary charge placed in a magnetic field is:
Answer: A. zero
Why: A magnetic force acts only on moving charges; a stationary charge feels none.
Q19.
A current-carrying solenoid behaves like a:
Answer: A. bar magnet
Why: A solenoid produces a field like that of a bar magnet, with a north and a south pole.
Q20.
The SI unit of magnetic flux is the:
Answer: A. weber
Why: Magnetic flux is measured in weber (Wb).
Q21.
A proton moves perpendicular to B=0.1 T field at 10<sup>6</sup> m/s. Radius of circular path (m<sub>p</sub> = 1.67 x 10<sup>-27</sup> kg, e = 1.6 x 10<sup>-19</sup> C):
Answer: A. 0.104 m
Why: r = mv/(qB) = 1.67x10<sup>-27</sup> x 10<sup>6</sup> / (1.6x10<sup>-19</sup> x 0.1) = 1.67x10<sup>-21</sup> / 1.6x10<sup>-20</sup> = 0.104 m.
Q22.
A long straight wire carries 10 A. Magnetic field at 0.1 m from wire (mu<sub>0</sub> = 4pi x 10<sup>-7</sup>):
Answer: A. 2 x 10<sup>-5</sup> T
Why: B = mu<sub>0</sub> I / (2 pi r) = 4pi x 10<sup>-7</sup> x 10 / (2 pi x 0.1) = 4x10<sup>-6</sup> / 0.2pi = 2x10<sup>-5</sup> T.
Q23.
For a solenoid of n turns per unit length carrying current I, internal field is:
Answer: B. mu<sub>0</sub> nI
Why: Solenoid: B = mu<sub>0</sub> nI (n = turns per unit length). This is uniform inside a long solenoid.
Q24.
Two parallel wires carry currents in same direction. They:
Answer: B. Attract each other
Why: Parallel currents in same direction attract (like charges repel but like currents attract). This is due to Ampere's force law.
Q25.
In a cyclotron, the frequency of applied alternating voltage (cyclotron frequency) is:
Answer: A. qB/(2pi m)
Why: Cyclotron frequency f = qB/(2 pi m). It depends on charge, mass, and field but NOT on speed (for non-relativistic particles).
Q26.
A charged particle in a uniform magnetic field moves in a:
Answer: C. Circle
Why: A charged particle moving perpendicular to a uniform magnetic field moves in a circle. The magnetic force provides centripetal force.
Q27.
Magnetic force on a current-carrying conductor in a field B is F = BIL sin(theta). Force is zero when:
Answer: C. theta = 0 degrees
Why: F = BIL sin(theta) = 0 when sin(theta) = 0, i.e., theta = 0 degrees. When current is parallel to B, no force.
Q28.
A current-carrying circular loop is placed in a uniform external magnetic field with its magnetic moment initially anti-parallel to the field. The loop is in:
Answer: B. Unstable equilibrium, since a small disturbance will cause it to rotate further away from this orientation
Why: When the magnetic moment is anti-parallel to the field, the potential energy U = -mB cos(θ) is at a maximum, so this is an unstable equilibrium and any small perturbation causes the loop to flip toward the stable, parallel orientation.
Q29.
A moving coil galvanometer can be converted into a voltmeter of higher range by:
Answer: C. Connecting a high resistance in series with the galvanometer coil
Why: A galvanometer is converted into a voltmeter by connecting a high resistance in series, which limits the current through the coil and extends the voltage range it can measure.
Q30.
A charged particle moves undeflected through a region containing both an electric field E and a perpendicular magnetic field B, with the fields oriented so the forces oppose each other. The speed of the particle must be:
Answer: D. v = E/B
Why: For the particle to move undeflected, the electric force qE must balance the magnetic force qvB, giving v = E/B; this principle is used in velocity selectors.
Q31.
The magnetic force F = qvB sinθ is maximum when the angle θ is:
Answer: A. 90°
Why: sin θ is greatest at 90°, so the force is maximum when v is perpendicular to B.
Q32.
The magnetic field at the centre of a circular loop of radius R carrying current I is:
Answer: A. μ₀I/2R
Why: At the centre of a circular loop, B = μ₀I/2R.
Q33.
The magnetic field due to a long straight wire at a perpendicular distance r is:
Answer: A. μ₀I/2πr
Why: For a long straight wire, B = μ₀I/2πr.
Q34.
A charged particle entering a uniform magnetic field perpendicular to it follows a path that is:
Answer: A. circular
Why: The constant perpendicular force provides centripetal force, giving a circular path.
Q35.
The radius of the circular path of a charge q of mass m moving at speed v in a field B is:
Answer: A. mv/qB
Why: Equating qvB = mv²/r gives r = mv/qB.
Q36.
Two long parallel wires carrying currents in the same direction:
Answer: A. attract each other
Why: Parallel currents attract; antiparallel currents repel.
Q37.
Ampère’s circuital law relates the magnetic field around a loop to the ___ enclosed by it:
Answer: A. current
Why: ∮B·dl = μ₀I<sub>enclosed</sub>, relating the field to the enclosed current.
Q38.
When a charged particle moves through a magnetic field, its ___ remains unchanged:
Answer: A. speed
Why: The magnetic force does no work, so it changes direction but not speed.
Q39.
The magnetic field inside a long solenoid with n turns per unit length carrying current I is:
Answer: A. μ₀nI
Why: For a long solenoid, B = μ₀nI, nearly uniform inside.
Q40.
A moving-coil galvanometer works on the principle that a current-carrying coil in a magnetic field experiences a:
Answer: A. torque
Why: The field exerts a torque on the coil, deflecting it in proportion to the current.
Q41.
A Hall probe measures charge carrier sign in a semiconductor. If holes are majority carriers, voltage measured is:
Answer: A. Positive on top (conventional current downward)
Why: Hall effect: positive holes drift and accumulate, creating Hall voltage. For p-type semiconductor, Hall voltage has different sign than n-type.
Q42.
Magnetic vector potential A is related to B by:
Answer: A. B = curl A (del x A)
Why: Magnetic field B = del x A (curl of the vector potential A). This is a fundamental relation in electromagnetism.
Q43.
Magnetic field at center of a current loop (radius R, current I):
Answer: B. mu<sub>0</sub> I/(2 R)
Why: B at center of circular loop = mu<sub>0</sub> I/(2R). This comes from Biot-Savart law integration around the full circle.
Q44.
A proton and an alpha particle enter the same perpendicular magnetic field with same speed. Ratio of radii r<sub>p</sub> : r<sub>alpha</sub>:
Answer: A. 1:2
Why: r = mv/(qB). m<sub>alpha</sub> = 4m<sub>p</sub>, q<sub>alpha</sub> = 2q<sub>p</sub>. r<sub>alpha</sub> = 4m<sub>p</sub> x v/(2q<sub>p</sub> B) = 2r<sub>p</sub>. Ratio r<sub>p</sub>:r<sub>alpha</sub> = 1:2.
Q45.
A velocity selector uses perpendicular E and B fields. Particles pass straight when:
Answer: A. v = E/B
Why: For straight-line motion: electric force qE = magnetic force qvB. So v = E/B (velocity selector condition).
Q46.
A circular coil of radius R carrying current I is placed with its plane perpendicular to a uniform magnetic field B. If the coil is now turned so its plane becomes parallel to B, the torque on the coil changes from its initial value to:
Answer: D. Maximum torque, since torque is greatest when the plane is parallel to the field
Why: Torque on a current loop is τ = mB sin(θ), where θ is the angle between the loop's normal and B; when the plane is parallel to B, the normal is perpendicular to B (θ=90°), giving maximum torque, opposite to the initial perpendicular-plane case where torque was zero.
Q47.
A toroid has a mean radius of 0.2 m and 500 turns, carrying a current of 4 A. The magnetic field inside the toroid (along the mean circumference) is approximately:
Answer: A. 4 × 10⁻³ T
Why: B = μ0 N I/(2πr) = (4π×10⁻⁷ × 500 × 4)/(2π × 0.2) ≈ 4 × 10⁻³ T.
Q48.
A current-carrying wire is bent into a semicircular arc of radius R, and current I flows through it. The magnetic field at the centre of the arc due to this semicircular section is:
Answer: B. μ0I/(4R)
Why: For a full circular loop, B at the centre is μ0I/(2R); a semicircular arc contributes exactly half of this, giving μ0I/(4R).
Q49.
An electron moving with speed v enters a region of uniform magnetic field B at an angle θ (not 90°) to the field, where 0° < θ < 90°. The path traced by the electron is:
Answer: C. A helix, since the velocity component along B continues unaffected while the perpendicular component causes circular motion
Why: The velocity component parallel to B experiences no magnetic force and continues unchanged, while the perpendicular component produces circular motion, and the combination traces a helical path.
Q50.
A long straight wire carrying current I<sub>1</sub> = 5 A is placed parallel to another wire carrying current I<sub>2</sub> = 10 A, 0.05 m apart, with currents in opposite directions. The force per unit length between the wires is approximately, and its nature is:
Answer: D. 2 × 10⁻⁴ N/m, repulsive
Why: Force per unit length = μ0 I<sub>1</sub> I<sub>2</sub>/(2π d) = (4π×10⁻⁷ × 5 × 10)/(2π × 0.05) = 2×10⁻⁴ N/m; currents in opposite directions repel each other.
Q51.
The time period T = 2πm/qB of a charged particle in a magnetic field is independent of the particle’s:
Answer: A. speed
Why: T depends on m, q and B but not on the speed, which is why a cyclotron works.
Q52.
Two parallel wires 1 m apart each carry 1 A. The force per unit length between them is:
Answer: A. 2 × 10⁻⁷ N/m
Why: F/L = μ₀I₁I₂/2πd = 2 × 10⁻⁷ × (1×1)/1 = 2 × 10⁻⁷ N/m.
Q53.
A proton and an electron enter the same magnetic field with equal speeds. The radius of the proton’s path is ___ the electron’s:
Answer: A. larger than
Why: Since r = mv/qB and the proton is far more massive, its radius is larger.
Q54.
In a cyclotron, the frequency of the alternating voltage is set equal to the ___ frequency of the particle:
Answer: A. cyclotron
Why: The applied frequency matches the cyclotron frequency qB/2πm so the particle is accelerated each half-cycle.
Q55.
The magnetic force acting on a moving charge can change its ___ but never its speed:
Answer: A. direction
Why: The force is always perpendicular to velocity, so it turns the particle without changing its speed or energy.
Q56.
A plane current loop of area A carrying current I has a magnetic moment equal to:
Answer: A. IA
Why: The magnetic dipole moment of a current loop is m = IA.
Q57.
The torque on a current loop of magnetic moment m in a uniform field B is:
Answer: A. mB sinθ
Why: τ = m × B = mB sinθ, where θ is the angle between m and B.
Q58.
To convert a galvanometer into an ammeter, a small resistance is connected:
Answer: A. in parallel, as a shunt
Why: A low-resistance shunt in parallel diverts most of the current, letting the meter read large currents.
Q59.
To convert a galvanometer into a voltmeter, a large resistance is connected:
Answer: A. in series with it
Why: A high resistance in series limits the current so the meter can read voltage across a large range.
Q60.
The work done by the magnetic force on a moving charged particle is always:
Answer: A. zero
Why: The force is perpendicular to the velocity, so it does no work on the charge.
Q61.
A charge of 1.6×10⁻¹⁹ C moves at 10⁶ m/s perpendicular to a magnetic field of 0.1 T. The magnetic force on it is:
Answer: B. 1.6×10⁻¹⁴ N
Why: F = qvB = 1.6×10⁻¹⁹·10⁶·0.1 = 1.6×10⁻¹⁴ N.
Q62.
A charged particle moves in a circle in a magnetic field. If its speed is doubled (field unchanged), the radius of its path:
Answer: B. doubles
Why: r = mv/qB, so r ∝ v; doubling the speed doubles the radius.
Q63.
A long solenoid has 1000 turns per metre and carries 2 A. The magnetic field inside it (μ₀ = 4π×10⁻⁷) is about:
Answer: B. 2.5×10⁻³ T
Why: B = μ₀nI = 4π×10⁻⁷·1000·2 ≈ 2.5×10⁻³ T.
Q64.
The cyclotron frequency of a charged particle in a magnetic field is:
Answer: A. independent of its speed
Why: f = qB/2πm depends only on charge, field and mass, not on speed or radius.
Q65.
The magnetic field at the centre of a circular loop of radius 0.1 m carrying 5 A (μ₀ = 4π×10⁻⁷) is about:
Answer: B. 3.14×10⁻⁵ T
Why: B = μ₀I/2R = 4π×10⁻⁷·5/(2·0.1) ≈ 3.14×10⁻⁵ T.
Q66.
A rectangular coil of 100 turns and area 0.01 m² carries 1 A in a field of 0.5 T with its plane parallel to the field. The torque on it is:
Answer: B. 0.5 N·m
Why: τ = NIAB = 100·1·0.01·0.5 = 0.5 N·m (maximum, since plane is parallel to field).
Q67.
In a velocity selector, the electric field is 1000 V/m and the magnetic field is 0.5 T. Undeflected particles have speed:
Answer: C. 2000 m/s
Why: v = E/B = 1000/0.5 = 2000 m/s.
Q68.
Two long parallel wires carry currents in the same direction. The wires:
Answer: A. attract each other
Why: Parallel currents in the same direction attract each other.