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Magnetism & Magnetic Circuits MCQ [Free PDF] – 13 Objective Questions with Answers

Every Magnetism & Magnetic Circuits question asked in RRB JE, GPSC DEE Electrical, GPSC AE Electrical, Supervisor Instructor, Vidyut Sahayak JE — solve, check instantly, then read the formula and exam tip.

13 PYQs11 subtopics5 examsUpdated 23 Oct 2026
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Q1Easy
What happens to the magnetic field around a straight current-carrying conductor when the current through it is increased?
View answer & solution
Correct answer: B. Magnetic field strength increases
Quick explanation

Around a long straight conductor B = μI / (2πr). At the same distance and in the same medium, B is directly proportional to I, so the field gets stronger.

B = μI / (2πr) At fixed r and μ: B ∝ I
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Q2Moderate
Which statement about magnetic flux is false?
View answer & solution
Correct answer: B. Magnetic flux does not have any SI unit.
Quick explanation

Magnetic flux Φ has an SI unit — the weber (Wb). Flux density B is in tesla (T) = Wb/m².

Φ in weber (Wb) B = Φ / A 1 T = 1 Wb/m²
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Mutual Inductance (2)

Q3Moderate
Two coils having self-inductances of 1 mH and 2 mH are mutually coupled. The maximum possible mutual inductance is ___________
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Correct answer: A. 1.414 mH
Quick explanation

Mutual inductance satisfies M = k√(L1L2), where 0 ≤ k ≤ 1. The maximum occurs at perfect coupling k = 1: Mmax = √(1×2) mH = √2 mH ≈ 1.414 mH.

M = k√(L₁L₂) Mmax = √(1×2) mH = 1.414 mH
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Q4Easy
If two coils have self inductance L₁ and L₂ , then mutual inductance M between them is given by:
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Correct answer: C. M = k√(L₁L₂)
Quick explanation

C — M = k√(L₁L₂). The coupling coefficient k accounts for the fraction of magnetic coupling between the two coils.

M = k√(L₁L₂) 0 ≤ k ≤ 1; M, L₁ and L₂ in H k = M/√(L₁L₂) M = k√(L₁L₂), with 0 ≤ k ≤ 1 √(L₁L₂) = √(4 × 9) = 6 H M = 0.5 × 6 = 3 H
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Coupled Coils in Series (1)

Q5Moderate
When two inductors L₁ and L₂ are connected in series opposition and M is the mutual inductance between them, the effective inductance L_eq is given by which equation?
View answer & solution
Correct answer: D. L = L₁ + L₂ − 2M
Quick explanation

D — L = L₁ + L₂ − 2M. In series opposition the mutually induced flux opposes the self-induced flux in each coil, so 2M is subtracted (series aiding gives +2M).

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Ferromagnetism (1)

Q6Easy
What is a material called that gets magnetised easily and retains its magnetisation for a long time?
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Correct answer: B. Ferromagnetic material
Quick explanation

B — Ferromagnetic material. In ferromagnetic materials (iron, cobalt, nickel) magnetic domains align strongly under an external field, so the material becomes strongly magnetised and can retain it.

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Inductance (1)

Q7Moderate
Find the inductance of a coil with permeability 3.5, turns 100 and length 2m. Assume the area to be thrice the length.
View answer & solution
Correct answer: A. 131.94 mH
Quick explanation

Inductance follows L = μN²A/l. Using the intended permeability/geometry data in the question gives approximately 131.94 mH, matching option A.

ℜ = l/(μA) L = N²/ℜ = μN²A/l Use μ = μ₀ μᵣ when the stated permeability is relative.
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Inductance of a Coil (1)

Q8Easy
If the length of a coil is increased, what is the effect on its inductance?
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Correct answer: D. Decreases
Quick explanation

D — Decreases. For a solenoid L = μN²A/l, so with other factors constant inductance is inversely proportional to length.

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Inductors in Series (1)

Kirchhoff Analogies (1)

Q10Easy
Can we apply Kirchhoff’s law to magnetic circuits?
View answer & solution
Correct answer: A. Yes
Quick explanation

Yes. Magnetic circuits use direct analogies of Kirchhoff's laws: flux continuity at a junction resembles KCL, and the algebraic sum of MMF drops around a magnetic path equals the applied MMF, analogous to KVL.

ΣΦ_in = ΣΦ_out NI = Σ(Φℜ) ℜ = l/(μA)
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MMF and Flux (1)

Q11Easy
What happens to the MMF when the magnetic flux decreases?
View answer & solution
Correct answer: B. Decreases
Quick explanation

For a magnetic circuit of unchanged reluctance, flux is proportional to magnetomotive force: Φ = F/ℜ. Therefore, if flux decreases, the MMF required/producing it also decreases.

F = NI ℜ = l/(μA) Φ = F/ℜ = NI/ℜ
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Reluctance and Permeability (1)

Q12Easy
Consider the following statements regarding magnetic circuits:
1. Magnetic reluctance represents the opposition offered by a material to the establishment of magnetic flux, analogous to resistance in electric circuits.
2. The permeability of a material is a constant parameter and does not vary with magnetic field intensity.
3. Introduction of an air gap in a magnetic circuit significantly increases the total reluctance due to low permeability of air.
4. Magnetic flux in a magnetic circuit is analogous to electric current flowing in an electrical circuit.
Which of the above statements is/are correct?
View answer & solution
Correct answer: A. 1, 3 and 4 only
Quick explanation

Magnetic circuits use an electric-circuit analogy: magnetomotive force drives magnetic flux through reluctance. An air gap greatly increases total reluctance because air has much lower permeability than iron. Key relation: ℜ=l/(μA); Φ=F/ℜ; F=NI.

Exam focus: Air gaps dominate reluctance even when physically short. Common trap: Permeability of magnetic materials is not generally constant with field intensity.

ℜ=l/(μA); Φ=F/ℜ; F=NI.
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Right-Hand Thumb Rule (1)

Q13Easy
The right-hand thumb rule is used to show the relationship between the direction of electric current and what?
View answer & solution
Correct answer: C. Magnetic field
Quick explanation

C — Magnetic field. For a straight current-carrying conductor, if the thumb points along the current, the curled fingers give the direction of the magnetic field lines.

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