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Force between parallel currents and definition of ampere: NEET MCQs with answers

49 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. Consider the following two statements regarding parallel current-carrying conductors: Statement I: Two parallel streams of protons travelling in vacuum in the same direction experience a mutually attractive magnetic force as well as a mutually repulsive electrostatic force. Statement II: The standard definition of one ampere historically adopted by SI is based on the magnetic interaction between two current-carrying parallel conductors of infinite length. 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
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Proton beams moving in the same direction constitute parallel currents, causing magnetic attraction, while their like positive charges cause electrostatic repulsion. Historically, the SI unit ampere was defined via the force of 2 × 10⁻⁷ N m⁻¹ between two infinitely long parallel conductors 1 m apart in vacuum.

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2. Two long parallel horizontal wires are held in the same vertical plane separated by a distance d. The lower wire has a mass per unit length λ and carries a current I₁. If it remains suspended in magnetic equilibrium against gravity due to the current I₂ in the upper wire, which condition must hold?

  1. The currents must flow in opposite directions, and I₁ I₂ = (4π d λ g) / μ₀
  2. The currents must flow in opposite directions, and I₁ I₂ = (2π d λ g) / μ₀
  3. The currents must flow in the same direction, and I₁ I₂ = (μ₀ d λ g) / (2π)
  4. The currents must flow in the same direction, and I₁ I₂ = (2π d λ g) / μ₀
Show answer and explanation

Answer: (D) The currents must flow in the same direction, and I₁ I₂ = (2π d λ g) / μ₀

For magnetic levitation, the upward magnetic force per unit length must balance the downward gravitational force (λ g). An upward force on the lower wire requires an attractive force towards the upper wire, meaning both currents must flow in the same direction. Balancing gives (μ₀ I₁ I₂) / (2π d) = λ g, which yields I₁ I₂ = (2π d λ g) / μ₀.

hard

3. Two parallel beams of electrons are moving in the same direction in vacuum with identical non-relativistic velocity v. What is the nature of the net force between the two electron beams?

  1. Always repulsive
  2. Always attractive
  3. Zero because electric and magnetic forces completely cancel
  4. Attractive at low speeds but repulsive at high speeds
Show answer and explanation

Answer: (A) Always repulsive

The electrons experience an electrostatic repulsive force F_e and a magnetic attractive force F_m due to parallel moving like charges. The ratio of magnetic force to electric force is F_m / F_e = v² / c². For non-relativistic speeds (v < c), F_m is strictly less than F_e, so the net interaction between the electron beams is always repulsive.

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4. Two infinitely long straight wires carrying currents I and 2I in the same direction are held parallel along the lines x = 0 and x = d respectively in the xy-plane. A third infinitely long straight wire carrying a current I₀ is to be placed parallel to both wires such that it experiences zero net magnetic force. The position of this third wire should be at:

  1. x = -d
  2. x = 2d / 3
  3. x = d / 2
  4. x = d / 3
Show answer and explanation

Answer: (D) x = d / 3

For the net magnetic force per unit length on the third wire to be zero, the net magnetic field produced by the two existing wires at its location must be zero. Since both currents are in the same direction, the zero-field point lies between them: B₁ = B₂ gives (μ₀ I)/(2π x) = (μ₀ (2I))/(2π (d - x)). Solving for x yields d - x = 2x, so 3x = d, giving x = d / 3.

easy

5. Two long parallel wires carry currents in the same direction. They:

  1. Rotate
  2. Attract each other
  3. Repel each other
  4. Exert no force on each other
Show answer and explanation

Answer: (B) Attract each other

Parallel currents attract and antiparallel currents repel, which is opposite to the behaviour of like electric charges.

NCERT Class 12

hard

6. Consider the following two statements regarding parallel current-carrying conductors: Statement I: Two stationary beams of electrons moving parallel to each other in vacuum in the same direction experience a net attractive force. Statement II: The magnetic force between two parallel currents satisfies Newton's third law of motion.

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

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

For two electron beams moving in the same direction, both electrostatic repulsion and magnetic attraction exist. Since the speed of electrons v < c, the electrostatic repulsion Fe = q²/(4πε₀r²) is strictly greater than the magnetic attraction Fm = (μ₀q²v²)/(4πr²), resulting in a net repulsive force. Statement II is correct as the mutual magnetic forces on two complete closed/infinite parallel circuits form an action-reaction pair.

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7. Three parallel coplanar long wires 1, 2, and 3 are arranged such that wire 2 is situated midway between wires 1 and 3. The separation between adjacent wires is 10 cm. The currents are I₁ = 5 A, I₂ = 3 A, and I₃ = 15 A, all flowing in the same direction. The net force per unit length on wire 2 is:

  1. 6 × 10⁻⁵ N m⁻¹ towards wire 1
  2. 3 × 10⁻⁵ N m⁻¹ towards wire 1
  3. 9 × 10⁻⁵ N m⁻¹ towards wire 3
  4. 6 × 10⁻⁵ N m⁻¹ towards wire 3
Show answer and explanation

Answer: (D) 6 × 10⁻⁵ N m⁻¹ towards wire 3

Wire 1 attracts wire 2 with f₁₂ = (2 × 10⁻⁷ × 5 × 3) / 0.10 = 3 × 10⁻⁵ N m⁻¹ (towards wire 1). Wire 3 attracts wire 2 with f₃₂ = (2 × 10⁻⁷ × 15 × 3) / 0.10 = 9 × 10⁻⁵ N m⁻¹ (towards wire 3). The net force is (9 - 3) × 10⁻⁵ = 6 × 10⁻⁵ N m⁻¹ directed towards wire 3.

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8. Two parallel, long conductors separated by a distance d carry currents I₁ and I₂ in the same direction. At what distance x from the conductor carrying current I₁ (in the region between them) is the net magnetic field zero?

  1. d I₁ / (I₂ - I₁)
  2. d I₂ / (I₁ + I₂)
  3. d I₁ / (I₁ + I₂)
  4. d √(I₁) / (√(I₁) + √(I₂))
Show answer and explanation

Answer: (C) d I₁ / (I₁ + I₂)

Between the two wires carrying currents in the same direction, the magnetic fields are opposite: B₁ = μ₀ I₁ / (2π x) and B₂ = μ₀ I₂ / [2π (d - x)]. Setting B₁ = B₂ gives I₁ / x = I₂ / (d - x), which simplifies to I₁(d - x) = I₂ x, hence x = d I₁ / (I₁ + I₂).

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9. Two long parallel wires carry unequal currents I₁ and I₂ such that I₁ = 3I₂. Wire 1 exerts a magnetic force of magnitude F₁ on wire 2, and wire 2 exerts a magnetic force of magnitude F₂ on wire 1. The ratio F₁ : F₂ is:

  1. 3 : 1
  2. 1 : 1
  3. 1 : 3
  4. 9 : 1
Show answer and explanation

Answer: (B) 1 : 1

According to Ampere's force law and Newton's third law, the force exerted by wire 1 on a given length of wire 2 is equal in magnitude and opposite in direction to the force exerted by wire 2 on the same length of wire 1. Therefore, F₁ / F₂ = 1 : 1, regardless of the unequal currents.

easy

10. In the international system of units (SI) prior to the 2019 redefinition, the standard ampere was defined based on the force between two parallel conductors. Which physical constant had an exactly defined numerical value without experimental uncertainty under that definition?

  1. Elementary charge e
  2. Permeability of free space μ₀
  3. Planck constant h
  4. Boltzmann constant k_B
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Answer: (B) Permeability of free space μ₀

Under the pre-2019 SI definition of the ampere, the magnetic force per unit length was fixed at exactly 2 × 10⁻⁷ N m⁻¹ for conductors separated by 1 m carrying 1 A. Consequently, the permeability of free space μ₀ was fixed exactly at 4π × 10⁻⁷ T m A⁻¹. In the 2019 revision, the ampere was redefined by fixing the elementary charge e.

hard

11. A straight wire carrying current I is placed parallel to the length of a flat, thin, conducting ribbon of width w carrying uniformly distributed current I in the same direction. The wire and ribbon lie in the same plane, and the distance between the wire and the nearest edge of the ribbon is d. The force per unit length between them is:

  1. (μ₀ I²)/(2π (d + w))
  2. (μ₀ I²)/(2π d) ln(1 + d/w)
  3. (μ₀ I²)/(2π w) ln(1 + w/d)
  4. (μ₀ I²)/(4π w) (w/d)
Show answer and explanation

Answer: (C) (μ₀ I²)/(2π w) ln(1 + w/d)

Consider a strip of width dx at distance x from the wire (from x = d to x = d + w). Current in strip is dI = (I/w) dx. Force per unit length on strip is dF/L = (μ₀ I dI)/(2π x) = (μ₀ I² dx)/(2π w x). Integrating from x = d to d + w yields F/L = (μ₀ I²)/(2π w) ln((d + w)/d) = (μ₀ I²)/(2π w) ln(1 + w/d).

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12. Two parallel, infinitely long straight conductors carrying currents I₁ = 10 A and I₂ = 30 A in opposite directions are separated by a distance of 10 cm in air. At what distance from wire 1 (carrying 10 A) on the line joining the two wires is the net magnetic field zero?

  1. 5 cm inside the region between the wires
  2. 5 cm outside the region between the wires, closer to wire 1
  3. 15 cm outside the region between the wires, closer to wire 2
  4. 20 cm outside the region between the wires, closer to wire 1
Show answer and explanation

Answer: (B) 5 cm outside the region between the wires, closer to wire 1

For opposite currents, the magnetic fields reinforce each other between the wires, so the null point cannot lie between them. It must lie outside the wires, closer to the weaker current (wire 1). Let x be the distance from wire 1 away from wire 2. Then B₁ = B₂ gives μ₀ I₁ / (2π x) = μ₀ I₂ / [2π (d + x)]. Substituting values: 10 / x = 30 / (10 + x), which gives 10 + x = 3x, so 2x = 10 cm and x = 5 cm outside wire 1.

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