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NEET Magnetic Effects of Current and Magnetism MCQs with answers

454 questions on Magnetic Effects of Current and Magnetism for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. A tightly wound solenoid of length 50 cm has 500 turns and carries a current of 2.5 A. What is the magnetic field near the centre on its axis? (Take μ₀ = 4π × 10⁻⁷ T·m/A)

  1. 3.14 × 10⁻³ T
  2. 1.57 × 10⁻³ T
  3. 6.28 × 10⁻³ T
  4. 7.85 × 10⁻⁴ T
Show answer and explanation

Answer: (A) 3.14 × 10⁻³ T

The number of turns per unit length is n = N / L = 500 / 0.50 m = 1000 m⁻¹. The magnetic field inside an ideal solenoid is B = μ₀ n I = (4π × 10⁻⁷) × 1000 × 2.5 = 10⁻³ π ≈ 3.14 × 10⁻³ T.

medium

2. Consider the following two statements regarding current-carrying conductors: Statement I: The net electric force between two current-carrying parallel copper wires is non-zero because mobile conduction electrons produce an external electrostatic field. Statement II: The magnetic force between two parallel currents is fundamentally relativistic in origin and is a consequence of Ampere's force law. In light of the above statements, choose the correct answer:

  1. Statement I is false but Statement II is true
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are true
  4. Both Statement I and Statement II are false
Show answer and explanation

Answer: (A) Statement I is false but Statement II is true

Statement I is false because an isolated current-carrying wire is electrically neutral overall (positive ion cores balance conduction electron charge), so the electrostatic force between them is zero. Statement II is true because the magnetic interaction between moving charges can be understood as a relativistic effect of electrostatics, and follows Ampere's force law.

hard

3. A rigid wire bent in the form of a semicircle of radius R lies in the xy-plane with its diameter on the x-axis from x = -R to x = +R. A current I flows through the semicircular arc from (-R, 0, 0) to (+R, 0, 0). The space has a uniform magnetic field B = B₀ (î + 2ĵ + 3k̂). The magnitude of the magnetic force on the wire is:

  1. 2 I R B₀ √10
  2. 2 I R B₀ √13
  3. 2 I R B₀ √5
  4. 4 I R B₀
Show answer and explanation

Answer: (B) 2 I R B₀ √13

The magnetic force on any shaped conductor in a uniform magnetic field depends only on the vector displacement from its start to end point: L_eff = (2R) î. The force is F = I (L_eff × B) = I [(2R î) × B₀(î + 2ĵ + 3k̂)] = 2 I R B₀ (î × î + 2 î × ĵ + 3 î × k̂) = 2 I R B₀ (2k̂ - 3ĵ). The magnitude of this force is |F| = 2 I R B₀ √(2² + (-3)²) = 2 I R B₀ √13.

medium

4. Two long, straight, parallel wires are separated by a distance of 20 cm and carry currents of 4 A and 6 A in opposite directions. The magnitude of the net magnetic field at a point midway between the two wires is:

  1. 1.0 × 10⁻⁵ T
  2. 2.0 × 10⁻⁵ T
  3. 4.0 × 10⁻⁵ T
  4. 0 T
Show answer and explanation

Answer: (B) 2.0 × 10⁻⁵ T

At the midpoint, r = 10 cm = 0.1 m from each wire. Because the currents flow in opposite directions, by the right-hand grip rule, both magnetic fields point in the same direction. B_net = B₁ + B₂ = (μ₀ / 2π r) (I₁ + I₂) = (2 × 10⁻⁷ / 0.1) × (4 + 6) = 2 × 10⁻⁶ × 10 = 2.0 × 10⁻⁵ T.

easy

5. A 0.2 m straight conductor carrying 8 A lies in a uniform 0.5 T magnetic field at 30° to the field. The force on it is:

  1. 1.6 N
  2. 20 N
  3. 0.4 N
  4. 0.8 N
Show answer and explanation

Answer: (C) 0.4 N

F = BIL sin θ = 0.5 × 8 × 0.2 × sin 30° = 0.4 N.

hard

6. A galvanometer of resistance 100 Ω gives full-scale deflection at 1 mA. The shunt resistance needed to convert it into an ammeter of range 0 to 1 A is:

  1. 1 Ω
  2. 0.1 Ω
  3. 99900 Ω
  4. 1 × 10⁵ Ω
Show answer and explanation

Answer: (B) 0.1 Ω

The shunt carries I − Ig: S = IgG/(I − Ig) = 0.001 × 100/0.999 = 0.1 Ω.

medium

7. A moving-coil galvanometer has a current sensitivity of 5 div/mA and a voltage sensitivity of 20 div/V. What is the resistance of the galvanometer coil?

  1. 400 Ω
  2. 250 Ω
  3. 100 Ω
  4. 50 Ω
Show answer and explanation

Answer: (B) 250 Ω

Current sensitivity is S_i = θ / I = 5 div/mA = 5 × 10³ div/A. Voltage sensitivity is S_v = θ / V = 20 div/V. Since V = I · G, we have S_v = S_i / G, which gives G = S_i / S_v = (5 × 10³) / 20 = 250 Ω.

medium

8. A galvanometer of resistance G requires a current of 5 mA for full-scale deflection. It is converted into a voltmeter of range 0 to 10 V by adding a resistance of 1950 Ω in series. If the same galvanometer is to be converted into an ammeter of range 0 to 2 A, what shunt resistance is needed?

  1. 0.050 Ω
  2. 0.250 Ω
  3. 0.125 Ω
  4. 0.500 Ω
Show answer and explanation

Answer: (C) 0.125 Ω

For voltmeter: V = I_g(G + R) ⇒ 10 = (5 × 10⁻³)(G + 1950) ⇒ G + 1950 = 2000 ⇒ G = 50 Ω. For ammeter of range 2 A: S = I_g × G / (I - I_g) = (0.005 × 50) / (2 - 0.005) = 0.25 / 1.995 ≈ 0.1253 Ω ≈ 0.125 Ω.

medium

9. A galvanometer of resistance 100 Ω gives full-scale deflection at 1 mA. The series resistance needed to convert it into a voltmeter of range 0 to 20 V is:

  1. 19900 Ω
  2. 0.005 Ω
  3. 20000 Ω
  4. 20100 Ω
Show answer and explanation

Answer: (A) 19900 Ω

At full scale the whole meter carries Ig, so G + R = V/Ig. R = V/Ig − G = 20/0.001 − 100 = 19900 Ω.

medium

10. An alpha particle (q = +2e, m = 4u) and a deuteron (q = +e, m = 2u) are accelerated from rest through the same potential difference V. They then enter perpendicularly into the same uniform magnetic field. The ratio of the radius of the alpha particle's path to that of the deuteron's path is:

  1. 1 : √2
  2. 1 : 1
  3. √2 : 1
  4. 1 : 2
Show answer and explanation

Answer: (B) 1 : 1

Radius R = (1/B) √(2m q V) / q = (1/B) √(2m V / q). Thus, R is proportional to √(m/q). For an alpha particle, m/q = 4/2 = 2. For a deuteron, m/q = 2/1 = 2. Therefore, R_alpha / R_deuteron = √(2/2) = 1, giving a ratio of 1 : 1.

medium

11. A circular coil of 10 turns and radius 25 cm carries a current of 7 A. The magnetic field at its centre (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 5.6 × 10⁻⁵ T
  2. 1.76 × 10⁻⁴ T
  3. 3.52 × 10⁻⁴ T
  4. 0.0176 T
Show answer and explanation

Answer: (B) 1.76 × 10⁻⁴ T

B = μ₀NI/(2R) = 4π × 10⁻⁷ × 10 × 7/(2 × 0.25) = 1.76 × 10⁻⁴ T.

medium

12. Consider the following statements regarding magnetic materials: Statement I: A diamagnetic gas placed in a non-uniform magnetic field tends to move slowly from regions of stronger magnetic fields to regions of weaker magnetic fields. Statement II: The magnetic dipole moment of an atom of a diamagnetic substance is non-zero in the absence of an external magnetic field. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect.

Statement I is correct because diamagnetic substances are repelled by magnetic fields and move toward regions of weaker field. Statement II is incorrect because individual atoms of a diamagnetic material have paired electrons and thus zero net intrinsic magnetic dipole moment in the absence of an external field.

medium

13. A circular coil of 100 turns and radius 20 cm carries a current of 8 A. The magnetic field at its centre (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 0.251 T
  2. 8 × 10⁻⁴ T
  3. 0.00503 T
  4. 0.00251 T
Show answer and explanation

Answer: (D) 0.00251 T

B = μ₀NI/(2R) = 4π × 10⁻⁷ × 100 × 8/(2 × 0.2) = 0.00251 T.

medium

14. In an electrical circuit, a battery of EMF 6 V and negligible internal resistance is connected across a resistor of 500 Ω. A voltmeter of resistance 2000 Ω is used to measure the potential difference across the resistor. The percentage error in the measurement of voltage is:

  1. 20%
  2. 25%
  3. 16.7%
  4. 0%
Show answer and explanation

Answer: (D) 0%

Because the battery has negligible internal resistance and is directly connected across the 500 Ω resistor, the potential difference across the resistor remains exactly equal to the EMF of the battery (6 V), regardless of the voltmeter's resistance. Hence, the voltmeter reads 6 V, and the percentage error is 0%.

easy

15. A short bar magnet having a magnetic dipole moment of 0.48 J T⁻¹ is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, what is the work done in turning it from its most stable to its most unstable orientation?

  1. 0.072 J
  2. 0.144 J
  3. 0.036 J
  4. 0.288 J
Show answer and explanation

Answer: (B) 0.144 J

Stable equilibrium corresponds to θ₁ = 0° and unstable equilibrium corresponds to θ₂ = 180°. The work done is W = -MB(cos 180° - cos 0°) = 2MB. Substituting M = 0.48 J T⁻¹ and B = 0.15 T gives W = 2 × 0.48 × 0.15 = 0.144 J.

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