NEET Magnetism and Matter PYQs & Practice Questions

10 free NEET Physics questions on Magnetism and Matter, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Circular Motion · Earth's Magnetism · Magnetic Susceptibility and Permeability · Bar Magnet and Magnetic Field · Axial and Equatorial Fields of a Bar Magnet · Torque on a Magnetic Dipole · Gauss's Law in Magnetism · Diamagnetism, Paramagnetism and Ferromagnetism · Curie's Law · Potential Energy of a Magnetic Dipole · Hysteresis and Ferromagnetism · Magnetic Dipole Moment

1. Q1NEET 2026Magnetic moment of a coil

   A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14×10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil, are respectively: (μ0=4π×10⁻⁷ T m/A)

   • A. 2.5 A, 20 A m²
   • B. 2 A, 4 A m²
   • C. 2.5 A, 2 A m²
   • D. 2 A, 10 A m²
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   Answer: B — 2 A, 4 A m²

   B=μ0NI/2R gives I=2.5 A; m=NIA=100×2.5×π(0.05)²≈2 A m².

2. Q2NEET 2026Magnetic field variation inside a wire

   A long straight solid wire of circular cross-section of radius 'a' carries steady current I, uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor is:

   • A. B increases linearly for r<a then stays constant for r>a
   • B. B increases linearly for r<a, peaks at r=a, then decreases as 1/r for r>a
   • C. B is constant inside and decreases as 1/r² outside
   • D. B decreases linearly inside and increases outside
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   Answer: B — B increases linearly for r<a, peaks at r=a, then decreases as 1/r for r>a

   Inside: B∝r; outside: B∝1/r; peak at r=a — matches option 3.

3. Q3NEET 2024Oscillating Magnetic Needle - Magnetic Moment Calculation

   In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8×10⁻⁶ kg m². If the magnitude of magnetic moment of the needle is x×10⁻⁵ A m²; then the value of 'x' is: [Figure: magnetic needle NS in uniform field B]

   • A. 50 π²
   • B. 1280 π²
   • C. 5 π²
   • D. 128 π²
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   Answer: A — 50 π²

   T = 2π√(I/μB). Time period T = 5/20 = 1/4 s. So 1/4 = 2π√(9.8×10⁻⁶/(μ×0.049)). Solving, μ = 128π² × 10⁻⁴ = 1280π² × 10⁻⁵. So x = 1280π².

4. Q4NEET 2024Magnetic Susceptibility of Materials (Match List)

   Match List-I with List-II List-I (Material): (A) Diamagnetic (B) Ferromagnetic (C) Paramagnetic (D) Non-magnetic List-II (Susceptibility (χ)): (I) χ=0 (II) 0>χ≥-1 (III) χ>>1 (IV) 0<χ<ε (a small positive number) Choose the correct answer from the options given below:

   • A. A-III, B-II, C-I, D-IV
   • B. A-IV, B-III, C-II, D-I
   • C. A-II, B-III, C-IV, D-I
   • D. A-II, B-I, C-III, D-IV
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   Answer: B — A-IV, B-III, C-II, D-I

   Diamagnetic materials have small negative susceptibility: 0>χ≥-1 (II). Ferromagnetic materials have very large positive susceptibility: χ>>1 (III). Paramagnetic materials have small positive susceptibility: 0<χ<ε (IV). Non-magnetic materials have χ=0 (I). So A-II, B-III, C-IV, D-I.

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5. Q5NEET 2024Magnetic Moment of a Bent Bar Magnet

   An iron bar of length L has magnetic moment M. It is bent at the middle of its length such that the two arms make an angle 60° with each other. The magnetic moment of this new magnet is:

   • A. 2M
   • B. M/√3
   • C. M
   • D. M/2
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   Answer: C — M

   Original magnetic moment M1 = mL = M where m is pole strength per unit length. After bending, each half-arm has length L/2, so new magnetic moment M2 = m(L/2) = M/2 (the resultant for bending at the midpoint such that pole strength is scaled with the half-length, treating it as simply halved since the magnetic moment of each arm is m×L/2, and by vector geometry at 60° the resultant equals M/2 per the given solution).

6. Q6NEET 2023Net Magnetic Flux Through Closed Surface

   The net magnetic flux through any closed surface is:

   • A. Positive
   • B. Infinity
   • C. Negative
   • D. Zero
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   Answer: C — Negative

   Since magnetic monopoles do not exist, magnetic field lines always form closed loops, so the net magnetic flux through any closed surface is always zero (Gauss's law for magnetism: ∮B·dA = 0).

7. Q7NEET 2022Circular Motion

   Proton moving with velocity v ⊥ to field B. Radius of circular path is:

   • A. r=mv/qB
   • B. r=qB/mv
   • C. r=mv/B
   • D. r=qBv/m
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   Answer: —

   Centripetal=magnetic force: mv²/r=qvB. r=mv/qB. Larger mass, velocity → larger radius. Larger charge, field → smaller radius.

8. Q8NEET 2022Dimensional Analysis - Magnetic Permeability

   The dimensions [MLT^-2 A^-2] belong to the:

   • A. Magnetic flux
   • B. Self inductance
   • C. Magnetic permeability
   • D. Electric permittivity
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   Answer: B — Self inductance

   Magnetic permeability (mu0) has dimensional formula [MLT^-2A^-2], derived from the formula for force between two current-carrying wires or from B = mu0*H relationships.

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9. Q9NEET 2022Solenoid Magnetic Field

   A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is

   • A. 6.28 × 10⁻² T
   • B. 12.56 × 10⁻² T
   • C. 12.56 × 10⁻⁴ T
   • D. 6.28 × 10⁻⁴ T
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   Answer: A — 6.28 × 10⁻² T

   n = 100 turns/mm = 10⁵ turns/m. B = μ0nI = 4π×10⁻⁷×10⁵×1 = 12.56×10⁻² T.

10. Q10NEET 2021Force on moving electron near current-carrying wire

    An infinitely long straight conductor carries a current of 5 A. An electron is moving with a speed of 10⁵ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant.

    • A. 4×10⁻²⁰ N
    • B. 8π×10⁻²⁰ N
    • C. 4π×10⁻²⁰ N
    • D. 8×10⁻²⁰ N
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    Answer: C — 4π×10⁻²⁰ N

    B=μ0I/2πr=(4π×10⁻⁷×5)/(2π×0.2)=5×10⁻⁶ T. F=qvB=1.6×10⁻¹⁹×10⁵×5×10⁻⁶=8×10⁻²⁰ N.