Moving Charges and Magnetism MCQ Quiz - Objective Question with Answer for Moving Charges and Magnetism - Download Free PDF

Last updated on Jul 8, 2025

Moving charges and magnetism are interconnected phenomena in physics. When electric charges are in motion, they create magnetic fields, and magnetism can induce electric currents. MCQs on moving charges and magnetism cover topics such as magnetic fields, magnetic force on moving charges, electromagnetic induction, magnetic properties of materials, and applications of magnetism. These Moving Charges and Magnetism MCQs assess knowledge of magnetic field concepts, electromagnetic induction principles, and interactions between electric currents and magnetic fields. Check how much you know about this physics concept with Moving Charges and Magnetism MCQs given here.

Latest Moving Charges and Magnetism MCQ Objective Questions

Moving Charges and Magnetism Question 1:

A solid conducting sphere with a radius of R and total charge q rotates about its diametric axis with a constant angular velocity ω. What is the magnetic moment of the sphere?

  1. (2/3) q R² ω
  2. (1/3) q R² ω
  3. (1/5) q R² ω
  4. (2/5) q R² ω

Answer (Detailed Solution Below)

Option 3 : (1/5) q R² ω

Moving Charges and Magnetism Question 1 Detailed Solution

Calculation:

The charge of disc element is 

dq = ρ2π r2 sinθ dθ dr

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The time of spining is dt = 2π/ ω 

The charge will be I = dq / dt =  ρωr2 sinθ dθ dr

The magnetic moment will be m = ∫ I da= ∫ ρωr2 sinθ dθ dr π (r2 sinθ2)

⇒  ρπω ∫rsin3θ dθ dr 

⇒ m = QωR2 / 5                        (ρ = 4Q / (3πR3) )

Moving Charges and Magnetism Question 2:

A proton is moving with a constant velocity and passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields, respectively, then this region of space may have:

(a) E = 0, B = 0

(b) E ≠ 0, B = 0

(c) E = 0, B ≠ 0

(d) E ≠ 0, B ≠ 0

  1. (a), (c), (d)
  2. (a), (b), (c), (d)
  3. (a), (b), (d)
  4. (b), (c), (d)

Answer (Detailed Solution Below)

Option 1 : (a), (c), (d)

Moving Charges and Magnetism Question 2 Detailed Solution

Calculation:

If both E and B are zero, then both the electric force (Fe) and the magnetic force (Fm) will be zero. Hence, the proton’s velocity may remain constant. Therefore, option (a) is correct.

If E = 0 and B ≠ 0, but the velocity of the proton is parallel or antiparallel to the magnetic field, then both Fe and Fm will be zero. Hence, option (c) is also correct.

If E ≠ 0 and B ≠ 0, but the net force Fe + Fm = 0, the velocity may again remain constant. Therefore, option (d) is also correct.

Thus, (a), (c), and (d) are correct options.

Moving Charges and Magnetism Question 3:

A solenoid having area A and length 'l' is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is :

  1. \(\rm \frac{B^2Al}{\mu_0}\)
  2. \(\rm \frac{B^2Al}{2\mu_0}\)
  3. B2 Al
  4. \(\rm \frac{B^2Al}{4\mu_0}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{B^2Al}{4\mu_0}\)

Moving Charges and Magnetism Question 3 Detailed Solution

Calculation:

Energy density (U/V) = B2 / (2μrμ0)

μr = 2 ⇒ U/V = B2 / (4μ0)

Total volume V = Aℓ

⇒ Total magnetic energy U = (B2 / 4μ0) × Aℓ

Final Answer: B2Aℓ / 4μ0

Hence, the correct option is (4).

Moving Charges and Magnetism Question 4:

In a moving coil galvanometer, two moving coils M, and M; have the following particulars :

R1 = 5Ω, N1 = 15, A1 = 3.6 x 10-3 m2, B1 = 0.25 T

R2 = 7Ω, N[2 = 21, A2 = 1.8 x 10-3 m2, B2 = 0.50 T

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M1 and M2 ? 

  1. 1 : 1
  2. 1 : 4
  3. 1 : 3
  4. 1 : 2

Answer (Detailed Solution Below)

Option 1 : 1 : 1

Moving Charges and Magnetism Question 4 Detailed Solution

Calculation:

Voltage sensitivity = θ/V = (NAB) / (cR)

Ratio = (N1A1B1 / N2A2B2) × (R2 / R1)

= (15 × 3.6 × 0.25) / (21 × 1.8 × 0.5) × (7 / 5)

= (13.5) / (18.9) × (7 / 5)

= (1 / 1)

Answer: 1 : 1

Moving Charges and Magnetism Question 5:

Two long parallel wires carrying currents 8A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is ________ × 10–6 T. (Given : √2 = 1.4)

Answer (Detailed Solution Below) 68

Moving Charges and Magnetism Question 5 Detailed Solution

Calculation:

Given two parallel wires carrying currents i₁ and i₂, the magnetic fields at a point due to each wire are perpendicular to each other. The magnetic field due to each wire is given by:

B₁ = (μ₀ i₁) / (2πd)

B₂ = (μ₀ i₂) / (2πd)

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The net magnetic field is the vector sum of the individual fields, as they are perpendicular to each other:

Bnet = √(B₁² + B₂²)

⇒ Bnet = (μ₀ / 2πd) × √(i₁² + i₂²)

⇒ Bnet = (4π × 10⁻⁷) / (2π × (7/√2) × 10⁻²) × √(8² + 15²)

⇒ Bnet = 68 × 10⁻⁶ T

Final Answer: Bnet = 68 × 10⁻⁶ T

Top Moving Charges and Magnetism MCQ Objective Questions

A particle of charge e and mass m moves with a velocity v in a magnetic field B applied perpendicular to the motion of the particle. The radius r of its path in the field is _______

  1. \(\frac{mv}{Be}\)
  2. \(\frac{Be}{mv}\)
  3. \(\frac{ev}{Bm}\)
  4. \(\frac{Bv}{em}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{mv}{Be}\)

Moving Charges and Magnetism Question 6 Detailed Solution

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CONCEPT:

  • When moving through a magnetic field, the charged particle experiences a force.
  • When the direction of the velocity of the charged particle is perpendicular to the magnetic field:
    • Magnetic force is always perpendicular to velocity and the field by the Right-Hand Rule.
    • And the particle starts to follow a curved path.
    • The particle continuously follows this curved path until it forms a complete circle.
    • This magnetic force works as the centripetal force.
  • Centripetal force (FC) = Magnetic force (FB)

​⇒ qvB = mv2/R

​⇒ R = mv/qB

where q is the charge on the particle, v is the velocity of it, m is the mass of the particle, B is the magnetic field in space where it circles, and R is the radius of the circle in which it moves.

F1 J.K 3.8.20 Pallavi D20

EXPLANATION:

Given that particle has charge e; mass = m; and moves with a velocity v in a magnetic field B. So

  • Centripetal force (FC) = Magnetic force (FB)

​​⇒ qvB = mv2/R

\(\Rightarrow R=\frac{mv}{qB}\)

\(\Rightarrow r=\frac{mv}{Be}\)

So the correct answer is option 1.

Magnetic Field inside a solenoid is ________.

  1. increases from one end to another
  2. uniform
  3. varies from point to point
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : uniform

Moving Charges and Magnetism Question 7 Detailed Solution

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CONCEPT:

  • Solenoid: A cylindrical coil of many tightly wound turns of insulated wire with a general diameter of the coil smaller than its length is called a solenoid.

F1 P.Y Madhu 14.04.20 D 3

  • A magnetic field is produced around and within the solenoid.
  • The magnetic field within the solenoid is uniform and parallel to the axis of the solenoid.

The strength of the magnetic field in a solenoid is given by:-

\(B=\frac{{{\mu }_{0}}NI}{l}\)

Where, N = number of turns, = length of the solenoid,  l = current in the solenoid and μo = absolute permeability of air or vacuum.

EXPLANATION:

  • The magnetic field inside a solenoid is uniform. So option 2 is correct.

In Fleming's left rule, the middle finger represents ________________.

  1. Force
  2. Direction of magnetic field
  3. Direction of current flowing through the conductor
  4. Magnetic Flux

Answer (Detailed Solution Below)

Option 3 : Direction of current flowing through the conductor

Moving Charges and Magnetism Question 8 Detailed Solution

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CONCEPT:

  • Fleming Left-hand rule gives the force experienced by a charged particle moving in a magnetic field or a current-carrying wire placed in a magnetic field.
    • This rule was originated by John Ambrose Fleming.
    • It is used in an electric motor.
    • It states that "stretch the thumb, the forefinger, and the central finger of the left hand so that they are mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the central finger points in the direction of motion of charge, then the thumb points in the direction of force experienced by positively charged particles."

GATE EE Reported 51

EXPLANATION:

  • In Fleming's left rule, the middle finger represents the direction of current flowing through the conductor. So option 3 is correct.
  • The thumb represents the direction of the magnetic force.
  • The forefinger represents the direction of the magnetic field.

Who discovered the presence of electron in an atom?

  1. Chadwick
  2. Niels Bohr
  3. Rutherford
  4. J.J. Thomson

Answer (Detailed Solution Below)

Option 4 : J.J. Thomson

Moving Charges and Magnetism Question 9 Detailed Solution

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The correct answer is J.J. Thomson.

Key Points

  • J.J. Thomson:
    • Electron was discovered by J. J. Thomson in 1897. So, option 4 is correct.
    • An electron is a low-mass and negatively charged particle.
    • He won the Nobel Prize in 1906 “for his theoretical and experimental investigations on the conduction of electricity by gases”.

Additional Information

  • Niels Bohr :
    • In 1913, Niels Bohr proposed a theory for the hydrogen atom based on quantum theory that energy is transferred only in certain well-defined quantities.
    • He was awarded the Nobel prize in 1922 "for his services in the investigation of the structure of atoms and of the radiation emanating from them."
  • J Chadwick:
    • He was a British physicist
    • He was associated with the discovery of Neutron and also awarded the noble prize for the discovery of neutrons.
  • Rutherford:
    • Rutherford was awarded the 1908 Nobel Prize in Chemistry for his theory of atomic structure.
    • He discovered the nucleus of the atom in 1911.
    • He is known as the father of Nuclear Physics.

The magnetic field is the strongest at

  1. middle of the magnet.
  2. north pole.
  3. south pole.
  4. both poles.

Answer (Detailed Solution Below)

Option 4 : both poles.

Moving Charges and Magnetism Question 10 Detailed Solution

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The correct answer is Both Poles.   

 Important Points

  • The ancient Greeks were the first known to have used this mineral, which they called a magnet because of its ability to attract other pieces of the same material and also iron.
  • Englishman William Gilbert was the first to investigate the phenomenon of magnetism systematically using scientific methods.
  • The closer the lines, the stronger the magnetic field, so in bar magnet, the magnetic field closest to the poles. 
  • It is equally stronger at both the poles, the force in the middle of the magnet is weaker and halfway between the poles and the centre.
  • The magnetic field is a vector quantity that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials.
  • A moving charge in a magnetic field experiences a force perpendicular to its own velocity and to that of the magnetic field.
  • S.I unit of the magnetic field is Tesla.

The magnetic dipole moment of a revolving electron is ____________. (e is charge on an electron, v is orbital speed and r is radius of orbit)

  1. evr/2
  2. 2ev/r
  3. 2er/v
  4. 2rv/e

Answer (Detailed Solution Below)

Option 1 : evr/2

Moving Charges and Magnetism Question 11 Detailed Solution

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CONCEPT: 

  • If the circular loop is considered as a magnetic dipole, then the dipole moment is the product of current and area. 
  • Therefore, the magnitude of magnetic moment = area × current 

M = IA

Where M = Magnetic moment of copper coil, I = Current flowing through a loop and A = Area of coil

EXPLANATION:

  • If e is the charge on an electron revolving in an orbit of radius r with uniform angular velocity ω, then equivalent current is

\(\Rightarrow I = \frac{charge (e)}{time (T)}\)

Where T = the period of revolution of electron

\(\Rightarrow T=\frac{2\pi r}{v}\)

∴ The equivalent  current will be 

\(\Rightarrow I = \frac{ev}{2\pi r}\)

  • Area of the orbit is

\(\Rightarrow A= \pi r^2\)

  • The magnitude of the magnetic moment

\(\Rightarrow M = \frac{ev}{2\pi r}\times\pi r^2=\frac{evr}{2}\)

Multiply and divide the above equation by m, we get

\(\Rightarrow M=\frac{emvr}{2m} = \frac{evr}{2}\)            

The value of the magnetic force is ________ in the center of the bar magnet.

  1. Maximum
  2. Minimum
  3. Zero
  4. Equal

Answer (Detailed Solution Below)

Option 3 : Zero

Moving Charges and Magnetism Question 12 Detailed Solution

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The correct option is Zero.

Key PointsBar magnet:

  • An object that is made up of ferromagnetic material, such as iron, steel, or a ferromagnetic composite, and permanently magnetic. Mostly bar magnets are rectangular or cylindrical in shape.
  • The magnetic field at the center of a bar magnet is zero.

Additional InformationProperties of Bar Magnet:

  • It has a north pole and a south pole at two ends. It doesn't matter how many pieces you break apart from a bar magnet, both pieces will still have a north pole and a south pole.
  • An imaginary line can be drawn along the magnetic field acting around any magnetic substance to define the magnetic field lines.
  • At the poles, the magnetic force is strongest.
  • If you suspend the bar magnet freely in the air it will align in a north-south direction. This property is used in the compass.
  • Unlike poles of a magnet that attract each other poles will repeal each other when we bring them close.
  • A ferromagnetic material is attracted by magnets like iron, cobalt, etc.

If the index finger points towards the north and the middle finger towards the east, when using Fleming's left-hand rule, what will be the direction of motion or that of the force acting on the conductor?

  1. South
  2. West
  3. Top 
  4. Bottom

Answer (Detailed Solution Below)

Option 3 : Top 

Moving Charges and Magnetism Question 13 Detailed Solution

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CONCEPT:

Fleming's Left-hand rule 

  • It gives the force experienced by a charged particle moving in a magnetic field or a current-carrying wire placed in a magnetic field.
  • It states that "stretch the thumb, the forefinger, and the central finger of the left hand so that they are mutually perpendicular to each other.
  • If the forefinger points in the direction of the magnetic field, the central finger points in the direction of motion of charge, then the thumb points in the direction of force experienced by positively charged particles."

GATE EE Reported 51

EXPLANATION:

According to question

  1. Forefinger (Index finger): In the direction north 
  2. Middle finger: In the direction east
  3. The thumb is pointing out of the paper i.e., on top.

The force acting on the conductor will be out of the paper i.e., on top. Therefore option 3 is correct.

Two solenoids having lengths L and 2L and the number of loops N and 4N, both have the same current, then the ratio of the magnetic field will be

  1. 1 : 2
  2. 2 : 1
  3. 1 : 4
  4. 4 : 1

Answer (Detailed Solution Below)

Option 1 : 1 : 2

Moving Charges and Magnetism Question 14 Detailed Solution

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CONCEPT:

  • solenoid is an instrument that consists of copper coiling over a cylinder designed to create a strong magnetic field inside the coil because of the flow of current through the coil.
  • By wrapping the same wire many times around a cylinder, the magnetic field due to the flow of current can become quite strong.
  • Hence we can say that the strength of the magnetic field will change as a current through coil or number of turn’s changes.
  • Magnetic field strength is independent of the diameter of the cylinder of a solenoid.
  • The strength of the magnetic field in a solenoid will be directly proportional to the number of turns and amount of current flowing through a wire and will be inversely proportional to its length.
  • Thus it is given by \(B = \frac{{{μ _0}NI}}{l}\)

Where N = number of turns and I = current, l = length of the solenoid

CALCULATION:
Given - Length of 1st solenoid = L, Length of 2nd solenoid = 2L, number of turns of 1st solenoid = N and number of turns of 2nd solenoid = 4L
  • Magnetic field due to the solenoid is given by
\(\Rightarrow B = \frac{{{μ _0}NI}}{l}\)
  • As μo and current (I) is constant, therefore 

\(\Rightarrow \;B \propto \frac{N}{L}\;\)

\( \Rightarrow \;\frac{{{B_1}}}{{{B_2}}}\; = \;\frac{{{N_1}}}{{{N_2}}} \times \frac{{{L_2}}}{{{L_1}}}\; = \;\frac{N}{{4N}} \times \frac{{2l}}{L}\; = \;\frac{1}{2}\)

The magnetic dipole moment of a current loop is independent of

  1. number of turns
  2. area of the loop
  3. current in the loop
  4. magnetic field in which it is lying

Answer (Detailed Solution Below)

Option 4 : magnetic field in which it is lying

Moving Charges and Magnetism Question 15 Detailed Solution

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CONCEPT:

  • When a circular loop is associated with the current I, it starts to act as a magnet and its magnetic moment is find as given below.

F1 J.K Madhu 10.07.20 D2

  • Magnetic moment (μ): The magnetic strength and orientation of a magnet or other object that produces a magnetic field.
    • It is a vector quantity associated with the magnetic properties of electric current loops.
    • It is equal to the amount of current flowing through the loop multiplied by the area encompassed by the loop.

μ = N i A 

where μ is the magnetic moment, A is the area of the coil, N is no. of turns and I is current in the coil.

  • Its direction is established by the right-hand rule for rotations.

EXPLANATION:

The Magnetic moment μ = N i A is independent of the magnetic field in which it is lying. 

  • As there is no magnetic field component in the formula.
  • So the correct answer is option 4.
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