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Explanation: The SI unit of magnetic field strength (or magnetic flux density) is Tesla (T), where 1 Tesla = 1 Newton / (Ampere · meter).
2. Oersted's experiment demonstrated that:
A. Changing magnetic fields induce electric current
B. An electric current produces a magnetic field around it
C. Magnetic monopoles exist
D. Light is an electromagnetic wave
Answer: B. An electric current produces a magnetic field around it
Explanation: Hans Christian Oersted discovered in 1820 that a magnetic needle deflects near a current-carrying wire, proving electric current creates a magnetic field.
3. A charge 'q' moving with velocity 'v' in a magnetic field 'B' experiences a Lorentz magnetic force given by:
A. F = q (v × B)
B. F = q (v · B)
C. F = (v × B) / q
D. F = q v B cosθ
Answer: A. F = q (v × B)
Explanation: Magnetic force on a moving charge is a vector cross product: F = q (v × B), with magnitude F = q v B sinθ.
4. What is the magnetic force on a charge moving parallel (θ = 0°) to the direction of a magnetic field?
A. Maximum (q v B)
B. Zero
C. q v B / 2
D. Infinite
Answer: B. Zero
Explanation: Since F = q v B sinθ and θ = 0°, sin 0° = 0, no magnetic force acts on a charge moving parallel or antiparallel to the field.
5. The path of a charged particle projected perpendicularly (θ = 90°) into a uniform magnetic field is a:
A. Straight line
B. Circle
C. Helix
D. Parabola
Answer: B. Circle
Explanation: When v is perpendicular to B, the magnetic force acts as a centripetal force (F = q v B = m v² / r), forcing the particle into a circular path.
6. If a charged particle enters a uniform magnetic field at an arbitrary angle θ (other than 0°, 90°, 180°), its path is:
A. Circular
B. Helical
C. Elliptical
D. Straight line
Answer: B. Helical
Explanation: The velocity component parallel to field causes linear motion while perpendicular component causes circular motion, producing a helical path.
7. Work done by a magnetic field on a moving charged particle is always:
A. Positive
B. Negative
C. Zero
D. Dependent on charge magnitude
Answer: C. Zero
Explanation: The magnetic force F is always perpendicular to velocity v (displacement). Work done W = F · d = 0, so magnetic fields do no work and kinetic energy remains constant.
8. According to Biot-Savart Law, the magnetic field dB due to a current element I dl at distance r varies as:
A. dB ∝ 1/r
B. dB ∝ 1/r²
C. dB ∝ r²
D. dB ∝ 1/r³
Answer: B. dB ∝ 1/r²
Explanation: Biot-Savart Law states dB = (μ₀ / 4π) * (I dl sinθ / r²), obeying an inverse-square distance dependence.
9. What is the value of permeability of free space (μ₀)?
A. 4π × 10^-7 T m A^-1
B. 9 × 10^9 N m² C^-2
C. 8.85 × 10^-12 C² N^-1 m^-2
D. 1.6 × 10^-19 C
Answer: A. 4π × 10^-7 T m A^-1
Explanation: The magnetic permeability constant of free space μ₀ is exactly 4π × 10^-7 T m A^-1 (or N/A²).
10. The magnetic field at the center of a circular current-carrying loop of radius R and N turns is:
A. B = μ₀ N I / (2 R)
B. B = μ₀ N I / (4π R)
C. B = μ₀ N I / (2π R)
D. B = μ₀ N I / R²
Answer: A. B = μ₀ N I / (2 R)
Explanation: Using Biot-Savart Law, magnetic field at center of a circular loop is B = μ₀ I / (2R); for N turns, B = μ₀ N I / (2R).
11. Ampere’s Circuital Law is expressed mathematically as:
A. ∮ B · dl = μ₀ I_enclosed
B. ∮ E · dl = -dΦ/dt
C. ∮ B · dA = 0
D. ∮ E · dA = Q/ε₀
Answer: A. ∮ B · dl = μ₀ I_enclosed
Explanation: Ampere's Law states line integral of magnetic field along any closed loop equals μ₀ times net current passing through loop area.
12. Magnetic field at distance 'r' from a long straight current-carrying wire carrying current 'I' is:
A. B = μ₀ I / (2π r)
B. B = μ₀ I / (4π r)
C. B = μ₀ I / (2 r)
D. B = 2 μ₀ I / r
Answer: A. B = μ₀ I / (2π r)
Explanation: Applying Ampere's Law around a circular path gives B(2πr) = μ₀ I, hence B = μ₀ I / (2π r).
13. Magnetic field inside a long straight ideal solenoid with 'n' turns per unit length carrying current 'I' is:
A. B = μ₀ n I
B. B = μ₀ N I / (2 L)
C. B = 0
D. B = μ₀ n I / 2
Answer: A. B = μ₀ n I
Explanation: Magnetic field inside a long tight solenoid is uniform and given by B = μ₀ n I, where n = N/L is turns per unit length.
14. What is the magnetic field outside an ideal long solenoid?
A. Maximum
B. Zero
C. μ₀ n I
D. Infinite
Answer: B. Zero
Explanation: For an ideal, infinitely long solenoid, field lines cancel outside, resulting in negligible (zero) magnetic field.
15. Force per unit length between two parallel straight current-carrying conductors separated by distance 'd' carrying currents I₁ and I₂ is:
A. F/L = (μ₀ I₁ I₂) / (2π d)
B. F/L = (μ₀ I₁ I₂) / (4π d²)
C. F/L = (μ₀ I₁ I₂) / (2 d)
D. F/L = μ₀ (I₁ + I₂) / d
Answer: A. F/L = (μ₀ I₁ I₂) / (2π d)
Explanation: Force per unit length between two long parallel wires is given by F/L = (μ₀ I₁ I₂) / (2π d).
16. Two parallel wires carrying current in the SAME direction will:
A. Repel each other
B. Attract each other
C. Neither attract nor repel
D. Rotate around each other
Answer: B. Attract each other
Explanation: Like currents (in same direction) attract each other, whereas opposite (antiparallel) currents repel each other due to magnetic forces.
17. Torque (τ) experienced by a current loop of magnetic dipole moment 'M' in a uniform magnetic field 'B' is:
A. τ = M × B
B. τ = M · B
C. τ = B × M
D. τ = M / B
Answer: A. τ = M × B
Explanation: Torque on a magnetic dipole in a magnetic field is τ = M × B, with magnitude τ = M B sinθ.
18. Magnetic dipole moment (M) of a current loop of N turns, carrying current I and enclosing area A is:
A. M = N I A
B. M = N I / A
C. M = I A / N
D. M = N A / I
Answer: A. M = N I A
Explanation: Magnetic moment of planar loop is vector product M = N I A, with direction given by right-hand thumb rule.
19. What is the gyromagnetic ratio of an electron revolving in an atom?
A. e / (2 m_e)
B. e / m_e
C. 2 e / m_e
D. e m_e / 2
Answer: A. e / (2 m_e)
Explanation: Gyromagnetic ratio is ratio of orbital magnetic moment to orbital angular momentum: μ_l / L = e / (2 m_e) ≈ 8.8 × 10^10 C/kg.
20. What is a Moving Coil Galvanometer used to measure or detect?
A. High voltage
B. Small electric currents
C. Resistance only
D. Magnetic field directly
Answer: B. Small electric currents
Explanation: Moving coil galvanometer is a sensitive instrument designed to detect and measure small currents in a circuit.
21. To convert a galvanometer into an ammeter, we connect:
A. Low resistance (shunt) in parallel
B. High resistance in series
C. High resistance in parallel
D. Low resistance in series
Answer: A. Low resistance (shunt) in parallel
Explanation: Connecting a low resistance (shunt) in parallel allows most current to bypass coil, converting galvanometer to ammeter.
22. To convert a galvanometer into a voltmeter, we connect:
A. High resistance in series
B. Low resistance in parallel
C. Low resistance in series
D. High resistance in parallel
Answer: A. High resistance in series
Explanation: A high resistance connected in series limits current through galvanometer when connected across potential difference, acting as a voltmeter.
23. What is the resistance of an ideal ammeter and an ideal voltmeter respectively?
A. Zero and Infinite
B. Infinite and Zero
C. Zero and Zero
D. Infinite and Infinite
Answer: A. Zero and Infinite
Explanation: Ideal ammeter has zero resistance to avoid altering circuit current; ideal voltmeter has infinite resistance so it draws no current.
24. What is current sensitivity of a galvanometer?
A. Deflection per unit current (θ / I = NBA / k)
B. Current per unit deflection
C. Deflection per unit voltage
D. Resistance per unit deflection
Answer: A. Deflection per unit current (θ / I = NBA / k)
Explanation: Current sensitivity is deflection produced per unit current: I_s = θ / I = (N B A) / k, where k is torsional constant.
25. In a Velocity Selector, charged particles pass undeflected when electric (E) and magnetic (B) fields satisfy:
A. v = E / B
B. v = B / E
C. v = E × B
D. v = E² / B
Answer: A. v = E / B
Explanation: Electric force qE equals magnetic force qvB when fields are mutually perpendicular and velocity v = E / B, allowing particle to pass un-deflected.
This page contains a chapter-wise objective practice set. Try the questions without checking the answer first, then use the explanation to understand the point you missed. The current page contains 25 questions, so you can practise in smaller sets when you are revising.
Quick revision tip
After the practice set, write down the question numbers you missed and revise only those concepts once more. This makes MCQ practice more useful than repeatedly attempting the same questions without reviewing mistakes.
