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Torque on magnetic dipoles; para-, dia- and ferromagnetism: NEET MCQs with answers

50 questions on this topic of Magnetic Effects of Current and Magnetism (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A magnetic dipole of magnetic moment M is placed in a uniform magnetic field B. The work done by an external agent in turning the dipole from an angle of 30° to 90° with respect to the direction of the magnetic field is:

  1. M B
  2. (1 / 2) M B
  3. [1 - (√3 / 2)] M B
  4. (√3 / 2) M B
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Answer: (D) (√3 / 2) M B

The work done by an external agent is W = -M B (cos θ₂ - cos θ₁) = M B (cos 30° - cos 90°) = M B [(√3 / 2) - 0] = (√3 / 2) M B.

medium

2. Read the following two statements regarding magnetic properties of materials: Statement I: The magnetic susceptibility of a diamagnetic substance is small, negative and independent of temperature. Statement II: Bismuth and copper show diamagnetism because their constituent atoms possess permanent non-zero magnetic moments. Choose the correct option:

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

Statement I is correct: diamagnetic susceptibility is negative, small and temperature-independent. Statement II is incorrect: atoms/molecules of diamagnetic substances have zero net permanent magnetic dipole moments due to paired electrons.

hard

3. A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles, each having a magnetic moment of 1.5 × 10⁻²³ A m². The sample is placed in a magnetic field of 0.8 T and cooled to a temperature of 4 K, reaching 15% of its saturation magnetization. What will be the total dipole moment of the sample if the field is changed to 0.4 T and temperature is increased to 16 K?

  1. 0.56 A m²
  2. 1.125 A m²
  3. 2.25 A m²
  4. 4.50 A m²
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Answer: (A) 0.56 A m²

Saturation magnetization M_sat = N × m = (2.0 × 10²⁴) × (1.5 × 10⁻²³) = 30 A m². At 4 K and 0.8 T, M₁ = 0.15 × 30 = 4.5 A m². By Curie's law (valid when far below full saturation), magnetization M ∝ B / T. Thus, M₂ / M₁ = (B₂ / T₂) / (B₁ / T₁) = (0.4 / 16) / (0.8 / 4) = 0.025 / 0.2 = 1/8. Therefore, M₂ = 4.5 / 8 = 0.5625 A m² ≈ 0.56 A m².

medium

4. A paramagnetic liquid is taken in a U-tube such that the meniscus is in between the pole pieces of an electromagnet. When the magnet is turned on so that a non-uniform field is created between the poles, what will happen to the level of the liquid in that limb?

  1. It rises slightly because paramagnetic fluids are pulled towards regions of stronger field.
  2. It falls slightly because paramagnetic fluids are expelled by strong fields.
  3. It remains unchanged because magnetic forces do not act on bulk liquids.
  4. It oscillates continuously about its mean position without settling down.
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Answer: (A) It rises slightly because paramagnetic fluids are pulled towards regions of stronger field.

Paramagnetic materials develop a weak magnetization in the direction of the magnetizing field and experience a net force directed from regions of weaker magnetic field to stronger magnetic field. Therefore, the liquid level rises in the limb placed between the pole pieces where the field is stronger.

easy

5. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north pole dipping downwards at an angle of 60° to the horizontal. If the horizontal component of Earth's magnetic field at that place is 0.2 × 10⁻⁴ T, what is the total magnetic field of the Earth at this location?

  1. 0.1 × 10⁻⁴ T
  2. 0.35 × 10⁻⁴ T
  3. 0.4 × 10⁻⁴ T
  4. 0.23 × 10⁻⁴ T
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Answer: (C) 0.4 × 10⁻⁴ T

The angle of dip is δ = 60°. The horizontal component is related to the total field B_E by B_H = B_E cos δ. Therefore, B_E = B_H / cos 60° = (0.2 × 10⁻⁴ T) / 0.5 = 0.4 × 10⁻⁴ T.

hard

6. A small magnetic dipole of moment M is positioned at the origin O along the +x-direction. A second identical dipole of moment M is placed at the point (d, 0, 0) pointing along the +y-direction. What is the magnitude of the net torque acting on the second dipole due to the magnetic field of the first?

  1. μ₀ M² / (2π d³)
  2. μ₀ M² / (4π d³)
  3. 2 μ₀ M² / (π d³)
  4. Zero
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Answer: (A) μ₀ M² / (2π d³)

The first dipole lies at the origin with M₁ = M î. The point (d, 0, 0) lies along its axial line (end-on position). The magnetic field produced by the first dipole at this location is B₁ = (μ₀ / 4π) (2M / d³) î. The second dipole has moment M₂ = M ĵ. The torque on the second dipole is τ = M₂ × B₁ = (M ĵ) × [(μ₀ / 4π) (2M / d³) î] = - (μ₀ M² / 2π d³) k̂. Its magnitude is therefore μ₀ M² / (2π d³).

medium

7. Above its Curie temperature, a ferromagnetic material becomes:

  1. Diamagnetic
  2. More strongly ferromagnetic
  3. A superconductor
  4. Paramagnetic
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Answer: (D) Paramagnetic

Heating a ferromagnet above its Curie temperature destroys the domain alignment, and the material behaves as a paramagnet.

NCERT Class 12

medium

8. Two identical thin bar magnets, each of length l and magnetic pole strength m, are placed perpendicular to each other such that the north pole of one touches the south pole of the other. What is the net magnetic dipole moment of the combined system?

  1. m l
  2. 2 m l
  3. m l / √2
  4. √2 m l
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Answer: (D) √2 m l

Each magnet has a dipole moment of magnitude M = m l directed from its south pole to its north pole. Since the two magnets are perpendicular, their dipole moment vectors M₁ and M₂ are mutually orthogonal. The magnitude of the resultant dipole moment is M_net = √(M₁² + M₂²) = √(M² + M²) = √2 M = √2 m l.

hard

9. A magnetic needle of magnetic moment M is aligned parallel to an external uniform magnetic field B. What is the ratio of the torque acting on it when rotated by 30° to the work done against magnetic forces in rotating it through the same angle of 30° from its initial position?

  1. (2 + √3)
  2. 1 / (2 - √3)
  3. (1 + √3)
  4. 1 / (1 - √3)
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Answer: (A) (2 + √3)

Torque at θ = 30° is τ = M B sin 30° = M B / 2. Work done to rotate from 0° to 30° is W = -M B (cos 30° - cos 0°) = M B (1 - √3/2) = M B (2 - √3)/2. The ratio τ / W is (1/2) / [(2 - √3)/2] = 1 / (2 - √3) = 2 + √3.

easy

10. When a diamagnetic rod is suspended freely in a uniform horizontal magnetic field, it slowly comes to rest with its length aligned:

  1. perpendicular to the magnetic field
  2. parallel to the magnetic field
  3. at an angle of 45° to the field
  4. in any arbitrary random direction
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Answer: (A) perpendicular to the magnetic field

Diamagnetic substances are feebly repelled by magnetic fields and tend to move from stronger to weaker parts of the field. Therefore, a diamagnetic rod aligns itself perpendicular to the external magnetic field lines to minimize field penetration.

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11. The magnetic susceptibility of a paramagnetic substance at 300 K is 1.2 × 10⁻⁵. What will be its magnetic susceptibility at 400 K?

  1. 0.6 × 10⁻⁵
  2. 1.6 × 10⁻⁵
  3. 0.9 × 10⁻⁵
  4. 1.2 × 10⁻⁵
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Answer: (C) 0.9 × 10⁻⁵

According to Curie's law, χ ∝ 1/T. Thus, χ₂/χ₁ = T₁/T₂. Substituting the values gives χ₂ = 1.2 × 10⁻⁵ × (300/400) = 0.9 × 10⁻⁵.

medium

12. Two statements are given below: Assertion (A): Electromagnets are made of soft iron rather than steel. Reason (R): Soft iron has high retentivity and high coercivity compared to steel. Choose the correct option:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
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Answer: (C) (A) is true but (R) is false

Electromagnets require materials that magnetize strongly when current flows but demagnetize rapidly when the current is switched off. Hence, soft iron is chosen because it possesses high permeability, high saturation magnetization, and low coercivity (with relatively narrow hysteresis loss), whereas steel has high coercivity. Thus, (A) is true but (R) is false.

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