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Explanation: Magnetic flux is given by Φ_B = B · A cosθ. Its SI unit is Weber (Wb), where 1 Wb = 1 Tesla · m².
2. Magnetic flux is a:
A. Scalar quantity
B. Vector quantity
C. Dimensionless quantity
D. Tensor quantity
Answer: A. Scalar quantity
Explanation: Magnetic flux is defined as the dot product of magnetic field vector B and area vector A (Φ_B = B · A), making it a scalar quantity.
3. According to Faraday’s Law of Electromagnetic Induction, the magnitude of induced EMF is proportional to:
A. Total magnetic flux
B. Rate of change of magnetic flux
C. Resistance of the circuit
D. Area of the coil only
Answer: B. Rate of change of magnetic flux
Explanation: Faraday's law states that induced EMF in a closed circuit equals the negative rate of change of magnetic flux linked with it (ε = -dΦ/dt).
4. Lenz’s Law gives the direction of:
A. Induced EMF / Induced Current
B. Magnetic field intensity
C. Electrostatic force
D. Electric charge
Answer: A. Induced EMF / Induced Current
Explanation: Lenz's Law specifies that the direction of induced current is such that it opposes the change in magnetic flux that produces it.
5. Lenz’s Law is a direct consequence of the principle of conservation of:
A. Charge
B. Energy
C. Momentum
D. Mass
Answer: B. Energy
Explanation: Mechanical work done against opposing electromagnetic forces is converted into electrical energy, satisfying conservation of energy.
6. The motional EMF induced in a straight conductor of length 'l' moving with velocity 'v' perpendicular to uniform field 'B' is:
A. ε = B l v
B. ε = B l / v
C. ε = B v / l
D. ε = B² l v
Answer: A. ε = B l v
Explanation: When a conductor cuts magnetic field lines perpendicularly at speed v, the induced motional EMF across its ends is given by ε = B l v.
7. A rod of length 'L' rotates with angular speed 'ω' about one end in a uniform magnetic field 'B' perpendicular to plane of rotation. The induced EMF is:
A. ε = 1/2 B L² ω
B. ε = B L² ω
C. ε = 2 B L² ω
D. ε = 1/2 B L ω
Answer: A. ε = 1/2 B L² ω
Explanation: Integrating motional EMF (v = ωr) along the radius yields total induced potential difference ε = 1/2 B L² ω.
8. What are Eddy Currents?
A. Direct currents in a circuit
B. Circulating currents induced in bulk conductors placed in changing magnetic fields
C. Alternating currents in capacitors
D. Leakage currents through insulators
Answer: B. Circulating currents induced in bulk conductors placed in changing magnetic fields
Explanation: Eddy currents (or Foucault currents) are closed loops of induced current produced inside solid metal blocks when magnetic flux varies through them.
9. Eddy currents are minimized in transformer cores by using:
A. Solid copper core
B. Laminated soft iron core insulated with varnish
C. Solid steel core
D. Hollow aluminum core
Answer: B. Laminated soft iron core insulated with varnish
Explanation: Thin insulated sheets (laminations) break up large circulation paths, increasing electrical resistance and drastically reducing eddy current energy loss.
10. Which device does NOT use Eddy Currents in its working?
A. Induction furnace
B. Electromagnetic braking in trains
C. Electric power meter
D. Step-up Transformer
Answer: D. Step-up Transformer
Explanation: Induction furnaces, magnetic brakes, and energy meters utilize eddy currents, whereas transformers aim to minimize them to avoid efficiency losses.
11. What is the SI unit of Self-Inductance (L) and Mutual Inductance (M)?
A. Weber
B. Farad
C. Henry
D. Tesla
Answer: C. Henry
Explanation: Both self and mutual inductance are measured in Henry (H), where 1 Henry = 1 Volt · second / Ampere.
12. Self-inductance of a coil is also known as:
A. Electrical Inertia
B. Electrical Capacity
C. Electrical Resistance
D. Magnetic Susceptibility
Answer: A. Electrical Inertia
Explanation: Self-inductance opposes any growth or decay of current in a circuit, acting as the electrical equivalent of mechanical inertia (mass).
13. Self-inductance (L) of a long solenoid of length 'l', cross-sectional area 'A', and total turns 'N' is:
A. L = μ₀ N² A / l
B. L = μ₀ N A / l
C. L = μ₀ N² A l
D. L = μ₀ N A l²
Answer: A. L = μ₀ N² A / l
Explanation: Magnetic flux NΦ = N (μ₀ N I A / l) = L I, giving self-inductance L = μ₀ N² A / l.
14. Energy stored in an inductor of inductance 'L' carrying steady current 'I' is:
A. U = 1/2 L I²
B. U = L I²
C. U = 1/2 L² I
D. U = L / I²
Answer: A. U = 1/2 L I²
Explanation: Work done in building up current against self-induced EMF is stored as magnetic potential energy U = 1/2 L I².
15. The coupling coefficient 'k' between two coils of self-inductances L₁ and L₂ with mutual inductance M is:
A. k = M / √(L₁ L₂)
B. k = M √(L₁ L₂)
C. k = (L₁ L₂) / M
D. k = M² / (L₁ L₂)
Answer: A. k = M / √(L₁ L₂)
Explanation: Coupling factor k ranges from 0 to 1 depending on magnetic flux linkage, defined by formula M = k √(L₁ L₂).
16. What is the peak EMF (ε₀) generated by an AC Generator coil of N turns, area A, rotating at angular speed ω in field B?
A. ε₀ = N B A ω
B. ε₀ = N B A / ω
C. ε₀ = N B / (A ω)
D. ε₀ = 1/2 N B A ω
Answer: A. ε₀ = N B A ω
Explanation: Instantaneous EMF is ε = N B A ω sin(ωt). The maximum amplitude or peak value is ε₀ = N B A ω.
17. Total charge 'q' induced in a circuit of resistance 'R' when magnetic flux changes by ΔΦ is:
A. q = ΔΦ / R
B. q = ΔΦ × R
C. q = R / ΔΦ
D. q = ΔΦ / (R × t)
Answer: A. q = ΔΦ / R
Explanation: q = ∫ I dt = ∫ (1/R dΦ/dt) dt = ΔΦ / R. Total induced charge depends only on total flux change and resistance, independent of time.
18. When a magnetic north pole is moved towards a closed circular coil, the front face of coil develops:
A. North polarity
B. South polarity
C. No polarity
D. Alternating polarity
Answer: A. North polarity
Explanation: By Lenz's law, induced current opposes approaching North pole by creating a North pole (counter-clockwise current) on the facing side.
19. Two coaxial circular coils carry current in same direction. If distance between them is increased, current in both coils will:
A. Increase
B. Decrease
C. Remain unchanged
D. Drop to zero
Answer: A. Increase
Explanation: Moving coils apart decreases linked magnetic flux. By Lenz's law, induced currents will oppose flux reduction by increasing in original direction.
20. Dimensional formula of Self-Inductance (L) is:
A. [M L² T^-2 A^-2]
B. [M L T^-2 A^-1]
C. [M L² T^-1 A^-2]
D. [M L^0 T^-2 A^-1]
Answer: A. [M L² T^-2 A^-2]
Explanation: From U = 1/2 L I², L = 2U / I². Dimensions = [M L² T^-2] / [A²] = [M L² T^-2 A^-2].
21. If current through an inductor increases at constant rate, induced EMF across it is:
A. Constant
B. Zero
C. Increasing linearly
D. Decreasing exponentially
Answer: A. Constant
Explanation: Magnitude of self-induced EMF is ε = L (dI/dt). Since rate of change of current (dI/dt) is constant, induced EMF remains constant.
22. If number of turns per unit length in a solenoid is doubled, its self-inductance becomes:
A. 2 times
B. 4 times
C. Half
D. Unchanged
Answer: B. 4 times
Explanation: Self-inductance L = μ₀ n² A l is proportional to square of turns per unit length (L ∝ n²). Doubling n increases L by factor of 2² = 4.
23. A metallic sheet is placed in a time-varying magnetic field. The sheet becomes hot due to:
A. Seebeck effect
B. Joule heating from Eddy currents
C. Peltier effect
D. Hysteresis loss only
Answer: B. Joule heating from Eddy currents
Explanation: Varying magnetic field induces eddy currents in metal bulk, causing electrical energy dissipation as heat (I²R loss).
24. What is the value of induced EMF in a coil when magnetic flux through it remains constant at 10 Wb?
A. 10 V
B. 1 V
C. Zero
D. 100 V
Answer: C. Zero
Explanation: Induced EMF relies on change of magnetic flux (ε = -dΦ/dt). If flux is constant (dΦ/dt = 0), induced EMF is zero.
25. Working of an AC Generator is based on the principle of:
A. Electromagnetic Induction
B. Mutual Inductance only
C. Heating Effect of Current
D. Coulomb's Law
Answer: A. Electromagnetic Induction
Explanation: AC Generator converts mechanical energy into electrical energy using electromagnetic induction by rotating a armature coil inside a magnetic field.
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.
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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.
