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Biot-Savart and Ampere laws with standard applications: NEET MCQs with answers

90 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 long straight wire carries 5 A. The magnetic field 2 cm from it (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 3.14 × 10⁻⁴ T
  2. 5 × 10⁻⁷ T
  3. 2.5 × 10⁻⁵ T
  4. 5 × 10⁻⁵ T
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Answer: (D) 5 × 10⁻⁵ T

B = μ₀I/(2πr) = 2 × 10⁻⁷ × 5/0.02 = 5 × 10⁻⁵ T.

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2. A circular loop of radius R carries a current I. At what distance x along the axis from the centre of the loop is the magnetic field equal to 1/8 of its value at the centre?

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

The axial field is B(x) = μ₀ I R² / [2(R² + x²)^(3/2)] and the central field is B(0) = μ₀ I / (2R). The ratio is B(x)/B(0) = R³ / (R² + x²)^(3/2) = 1/8. Taking the 2/3 power gives (R² + x²) / R² = 4, so R² + x² = 4R², which means x² = 3R² or x = √3 R.

hard

3. Which of the following statements about Ampere's circuital law is NOT correct?

  1. Even if the net current enclosed by an Amperian loop is zero, the magnetic field at points on the loop need not be zero everywhere.
  2. The line integral of magnetic field ∮ B · dl along any closed loop is independent of the shape and size of the loop.
  3. The current I_enclosed in Ampere's law includes only the currents enclosed by the Amperian loop.
  4. Ampere's circuital law is not valid for conductors of non-symmetric cross-sections.
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Answer: (D) Ampere's circuital law is not valid for conductors of non-symmetric cross-sections.

Ampere's circuital law is universally valid for any steady current distribution regardless of symmetry, although it is most useful for calculating magnetic fields when high symmetry is present. Therefore, statement A is false and the correct choice.

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4. A circular coil of 10 turns and radius 20 cm carries a current of 2 A. The magnetic field at its centre (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 2 × 10⁻⁵ T
  2. 1.26 × 10⁻⁴ T
  3. 0.00628 T
  4. 6.28 × 10⁻⁵ T
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Answer: (D) 6.28 × 10⁻⁵ T

B = μ₀NI/(2R) = 4π × 10⁻⁷ × 10 × 2/(2 × 0.2) = 6.28 × 10⁻⁵ T.

easy

5. A long straight wire carries 12 A. The magnetic field 50 cm from it (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 4.8 × 10⁻⁸ T
  2. 3.02 × 10⁻⁵ T
  3. 2.4 × 10⁻⁶ T
  4. 4.8 × 10⁻⁶ T
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Answer: (D) 4.8 × 10⁻⁶ T

B = μ₀I/(2πr) = 2 × 10⁻⁷ × 12/0.5 = 4.8 × 10⁻⁶ T.

hard

6. A square wire frame of side length 2a carries a steady current I. What is the magnetic field at the centre of the square?

  1. μ₀ I / (2π a)
  2. 2√2 μ₀ I / (π a)
  3. √2 μ₀ I / (π a)
  4. μ₀ I / (4π a)
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Answer: (C) √2 μ₀ I / (π a)

The centre is at a distance d = a from each of the 4 sides. For one finite side of length 2a, the angles subtended at the centre are θ₁ = θ₂ = 45°. Field due to one side is B₁ = (μ₀ I / 4π a)(sin 45° + sin 45°) = (μ₀ I / 4π a)(2 / √2) = √2 μ₀ I / (4π a). For all 4 sides in the same direction, B_net = 4 B₁ = √2 μ₀ I / (π a).

easy

7. A wire is bent into a semicircle of radius R carrying a current I. The magnetic field at the centre of curvature due to this semicircular part alone is:

  1. μ₀ I / (2 R)
  2. μ₀ I / (4 R)
  3. μ₀ I / (8 R)
  4. μ₀ I / (4π R)
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Answer: (B) μ₀ I / (4 R)

The magnetic field at the centre of a complete circular coil is B = μ₀ I / (2R). For a semicircular segment, it contributes exactly half of the total field, which gives B = (1/2) · [μ₀ I / (2R)] = μ₀ I / (4R).

easy

8. A long straight wire carries 4 A. The magnetic field 50 cm from it (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 8 × 10⁻⁷ T
  2. 1.6 × 10⁻⁶ T
  3. 1.6 × 10⁻⁸ T
  4. 1.01 × 10⁻⁵ T
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Answer: (B) 1.6 × 10⁻⁶ T

B = μ₀I/(2πr) = 2 × 10⁻⁷ × 4/0.5 = 1.6 × 10⁻⁶ T.

easy

9. A wire carrying a current I is bent into the shape of a circular arc of radius R subtending an angle of 120° at its centre O. The magnetic field at the centre O due to this arc is:

  1. μ₀ I / (12 R)
  2. μ₀ I / (3 R)
  3. μ₀ I / (6 R)
  4. μ₀ I / (4 R)
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Answer: (C) μ₀ I / (6 R)

A full circle of radius R produces a field B = μ₀ I / (2R) at its centre. An arc subtending an angle θ produces a field B = (θ / 2π) × (μ₀ I / (2R)). For θ = 120° = 2π/3 radians, B = (1/3) × (μ₀ I / (2R)) = μ₀ I / (6R).

easy

10. A long straight wire carries 3 A. The magnetic field 50 cm from it (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 7.54 × 10⁻⁶ T
  2. 1.2 × 10⁻⁶ T
  3. 6 × 10⁻⁷ T
  4. 1.2 × 10⁻⁸ T
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Answer: (B) 1.2 × 10⁻⁶ T

B = μ₀I/(2πr) = 2 × 10⁻⁷ × 3/0.5 = 1.2 × 10⁻⁶ T.

easy

11. A long straight wire carries 20 A. The magnetic field 10 cm from it (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 2 × 10⁻⁵ T
  2. 2.51 × 10⁻⁴ T
  3. 4 × 10⁻⁷ T
  4. 4 × 10⁻⁵ T
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Answer: (D) 4 × 10⁻⁵ T

B = μ₀I/(2πr) = 2 × 10⁻⁷ × 20/0.1 = 4 × 10⁻⁵ T.

easy

12. A long straight wire carries 4 A. The magnetic field 10 cm from it (μ₀ = 4π × 10⁻⁷ T m A⁻¹) is:

  1. 8 × 10⁻⁸ T
  2. 4 × 10⁻⁶ T
  3. 8 × 10⁻⁶ T
  4. 5.03 × 10⁻⁵ T
Show answer and explanation

Answer: (C) 8 × 10⁻⁶ T

B = μ₀I/(2πr) = 2 × 10⁻⁷ × 4/0.1 = 8 × 10⁻⁶ T.

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