16 NEET previous-year questions on Electromagnetic Induction, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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Zero
Bv π r 2 /2 and P is at higher potential
π rBv and R is at higher potential
2rBv and R is at higher potential
Solution
The potential difference across the semicircular ring is given by . The length of the semicircular ring is , so . Since the motion is downward and the magnetic field is horizontal, the potential at R is higher due to the direction of the induced current. Therefore, the potential difference is and R is at higher potential, but the correct option is (d) and R is at higher potential, considering the full length of the semicircle.
0.15 Nm
0.20 Nm
0.24 Nm
0.12 Nm www.vedantu.com 39
Solution
τ⃗ = M⃗⃗⃗ × B⃗⃗ =MBsin60o =Ni ABsin60o 50 ×2×0.12 ×0.1 ×0.2 × √3 2 =12√3 ×10−2 Nm=0.20748 Nm
Cells
Potential gradients
A condition of no current flow through the galvanometer
A combination of cells, galvanometer and resistances
Solution
Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit.
2 mA
0.2 A
2 A
0 ampere
Solution
Sol – +ε C R3R2L1 R1 L2 At t = 0, no current flows through R 1 and R3 ∴ + –ε i R2 2 i R ε= = 18 9 = 2 A Note : Not correctly framed but the best option out of given is (3).
1·389 H
138·88 H
0·138 H
13·89 H ACHLA/AA/Page 5 SPACE FOR ROUGH WORK English
Solution
The magnetic potential energy stored in an inductor is given by . Substituting the values, , so option (d) is correct.
68 cm
6.8 cm
113.9 cm
88 cm
Solution
Δl1 = Δl2 l1∝1 Δθ = l2∝2Δθ 88 × 1.7 × 10−5 = l2 × 2.2 × 10−5 Δl2 = 88×1.7/2.2 = 68 cm
12.2 nm
12.2 × 10−13 m
12.2 × 10−12 m
12.2 × 10−4 nm
Solution
λ = ℎ √2𝑚𝐾= 6.63×10−34 √2×9.1×10−31×104×1.6×10−19 = 12.2 ×10−12 m
metals
insulators only
semiconductors only
insulators and semiconductors
Solution
Metals have a positive temperature coefficient of resistance, while insulators and semiconductors have a negative temperature coefficient of resistance. This is due to the increase in carrier concentration in insulators and semiconductors with temperature, so option (d) is correct.
1 2 R R
2 1 R R
2 1 2 R R
2 2 1 R R
Solution
The mutual inductance between two coaxial circular loops of radii and (with ) is directly proportional to . This is derived from the formula for mutual inductance of two coaxial loops, so option (b) is correct.
3 Ia 2 and 3 Ia 2
3 Ia 2 and Ia 2
3 Ia 2 and 4 Ia 2
4 Ia 2 and 3 Ia 2
Solution
For the equilateral triangle, the number of turns and the area . The magnetic dipole moment is . For the square, and , so . Therefore, the correct option is (b).
Magnetic flux
Self inductance
Magnetic permeability
Electric permittivity
Solution
Dimensional formula of magnetic permeability is [MLT–2A–2] - 6 - NEET (UG)-2022 (Code-Q1)
2 weber
0.5 weber
1 weber
Zero weber
Solution
Magnetic flux () ·φ= ρρ B BA and ρρ BA are in same direction, therefore φB = B.A = 0.5 × 12 = 0.5 Wb
8 μJ
4 μJ
4 mJ
8 mJ
Solution
The magnetic energy stored in an inductor is given by . Substituting the values, , so option (a) is correct.
AB and DC
BA and CD
AB and CD
BA and DC
Solution
North of magnet is moving away from solenoid 1 so end B of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end C of solenoid 2 is South.
1 : 2
2 : 1
1 : 1
1 : 4
Solution
According to transformer ratio, 2 :1SS PP VN VN ==
Zero at all places
Constant between the plates and zero outside the plates
Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates
Zero between the plates and non-zero outside
Solution
Let the surface charge density be
Given
It means displacement current is constant.
This system will act like a cylindrical wire.
The graph of magnetic field (B) vs r is
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