Electromagnetic Induction NEET PYQReal NEET & AIPMT 2006–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Electromagnetic Induction question from the NEET & AIPMT papers on this site (2006–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
- Chapter
- NCERT Class 12
- PYQs here
- 13
- Papers
- NEET & AIPMT 2006–2021
- In the app
- 199+ questions
Electromagnetic Induction previous year questions
13 questions. Tap an option to check your answer.
Q1NEET 2021
Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ >> R₂, the mutual inductance M between them will be directly proportional to :
Not quite — the answer is D.
A current I in the large loop gives B = μ₀I/(2R₁) at its centre, nearly uniform over the small loop. Flux through the small loop = (μ₀I/2R₁)πR₂², so M = μ₀πR₂²/(2R₁) ∝ R₂²/R₁.
Q2NEET 2019 · Q41
A 800 turn coil of effective area 0.05 m² is kept perpendicular to a magnetic field 5 × 10⁻⁵ T. When the plane of the coil is rotated by 90° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:
Not quite — the answer is D.
Flux changes from NBA to zero: Δφ = 800 × 5 × 10⁻⁵ × 0.05 = 2 × 10⁻³ Wb. emf = Δφ/Δt = 2 × 10⁻³/0.1 = 0.02 V. 2 × 10⁻³ V is the flux change, not divided by the time.
Q3NEET 2018 · Q21
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
Not quite — the answer is D.
U = ½LI², so L = 2U/I² = 2 × 25 × 10⁻³/(60 × 10⁻³)² = 0.05/0.0036 ≈ 13.89 H. Forgetting to square the current or the factor 2 gives the other values.
Q4NEET 2017 (Gujarati paper) · Q164
A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. The induced e.m.f. in the coil is: (Take H_E = 3.0 × 10⁻⁵ T)
Not quite — the answer is C.
A 180° turn reverses the flux, so ΔΦ = 2NBA = 2 × 500 × 3.0 × 10⁻⁵ × π(0.1)² ≈ 9.42 × 10⁻⁴ Wb. Average emf = ΔΦ/Δt = 9.42 × 10⁻⁴ / 0.25 ≈ 3.8 × 10⁻³ V. The 2 Ω resistance affects the current, not the emf; dropping the factor 2 would give about 1.9 × 10⁻³ V.
Q5NEET 2016 Phase 1 · Q42
A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is:
Not quite — the answer is D.
Total flux linkage NΦ = LI, so L = NΦ/I = 1000 × 4 × 10⁻³/4 = 1 H. Forgetting to divide by the current gives 4 H.
Q6NEET 2013
A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is -
Not quite — the answer is B.
The induced emf in a rotating loop is e = NBAω sin ωt, which changes sign twice in each full rotation (at 0 and π). So the direction reverses twice per revolution.
Q7AIPMT 2012 Prelims
A coil of resistance 400 Ω is placed in a magnetic field. If the magnetic flux φ (Wb) linked with the coil varies with time t (s) as φ = 50t² + 4, the current in the coil at t = 2 s is
Not quite — the answer is B.
Induced emf e = dφ/dt = 100t, which is 200 V at t = 2 s. Current = e/R = 200/400 = 0.5 A. The constant 4 in φ does not contribute to the emf.
Q8AIPMT 2010 Prelims · Q36
A conducting circular loop is placed in a uniform magnetic field, B = 0.025 T with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of 1 mm s⁻¹. The induced emf when the radius is 2 cm is
Not quite — the answer is C.
φ = Bπr², so |ε| = Bπ·2r·(dr/dt) = 0.025 × π × 2 × 0.02 × 10⁻³ = π × 10⁻⁶ V = π μV. Forgetting the factor 2 from d(r²)/dt gives (π/2) μV.
Q9AIPMT 2009 Prelims · Q23
A rectangular, a square, a circular and an elliptical loop, all in the (x−y) plane, are moving out of a uniform magnetic field with a constant velocity, v⃗ = v î. The magnetic field is directed along the negative z axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for:
Not quite — the answer is A.
Induced emf = Bvl, where l is the length of the loop lying on the field boundary as it leaves. For a rectangle or square moving perpendicular to a side, l is a constant side length, so the emf is constant; for a circle or ellipse the chord at the boundary changes continuously, so the emf changes. A shrinking flux alone does not make the emf vary; it is the rate of change of flux that matters.
Q10AIPMT 2009 Prelims · Q42
A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm/s. The induced emf in the loop when the radius is 2 cm is:
Not quite — the answer is D.
Emf = B dA/dt = B × 2πr × dr/dt = 0.04 × 2π × 0.02 × 2 × 10⁻³ = 3.2π × 10⁻⁶ V = 3.2π μV. Missing the factor 2 from d(πr²)/dt gives 1.6π μV.
Q11AIPMT 2008 Prelims · Q35
A circular disc of radius 0.2 metre is placed in a uniform magnetic field of induction (1/π) Wb/m² in such a way that its axis makes an angle of 60° with B⃗. The magnetic flux linked with the disc is
Not quite — the answer is B.
The disc's area vector lies along its axis, so φ = BA cos 60° = (1/π)(π × 0.2²)(1/2) = 0.02 Wb. Using sin 60° (treating the angle as being with the plane of the disc) would give about 0.035 Wb.
Q12AIPMT 2008 Prelims · Q38
A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 × 10⁻³ Wb. The self-inductance of the solenoid is
Not quite — the answer is D.
Total flux linkage Nφ = LI, so L = 500 × 4 × 10⁻³/2 = 1.0 H. 2.0 H would forget to divide by the current.
Q13AIPMT 2006 Prelims · Q24
Two coils of self-inductances 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is :
Not quite — the answer is C.
With all the flux of one coil linking the other (coupling coefficient 1), M = √(L₁L₂) = √(2 × 8) = 4 mH. Adding the inductances (10 mH) or taking the larger one (8 mH) is incorrect.
How often Electromagnetic Induction came up
Questions from this chapter in each paper on this site.
Keep going
PYQs show what NEET asks. For the full Electromagnetic Induction bank (199+ questions with solutions), use the RankUp app.
