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Force on moving charges and current-carrying conductors: NEET MCQs with answers

75 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 proton, a deuteron, and an alpha particle enter a uniform magnetic field with equal linear momentum perpendicular to the magnetic lines of force. The ratio of the radii of their circular paths (r_p : r_d : r_α) is:

  1. 1 : 1 : 2
  2. 2 : 2 : 1
  3. 1 : 2 : 4
  4. 2 : 1 : 1
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Answer: (B) 2 : 2 : 1

The radius of a charged particle in a magnetic field is r = p / (qB). Since momentum p and magnetic field B are the same for all three particles, r is inversely proportional to charge q. The charges of proton, deuteron, and alpha particle are e, e, and 2e respectively. Therefore, r_p : r_d : r_α = (1/1) : (1/1) : (1/2) = 2 : 2 : 1.

medium

2. A closed flexible circular wire loop of radius R carries a steady current I. It is placed in a uniform magnetic field B directed perpendicular to the plane of the loop pointing outwards. The tension developed in the wire loop due to the magnetic force is:

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

Consider a small arc of the loop subtending an angle dθ at the center. The length of the arc is dl = R dθ. The outward radial magnetic force on this element is dF = I (dl) B = I R dθ B. This force is balanced by the inward radial components of tension T at both ends: 2 T sin(dθ/2) ≈ T dθ. Equating the two gives T dθ = I R B dθ, which simplifies to T = I R B.

hard

3. A straight wire of length 0.5 m and mass 50 g carries a current I. It is suspended in mid-air in horizontal equilibrium by a uniform horizontal magnetic field of 0.4 T perpendicular to the wire. Taking g = 9.8 m s⁻², the magnitude of the current I required is:

  1. 1.25 A
  2. 2.45 A
  3. 4.90 A
  4. 0.49 A
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Answer: (B) 2.45 A

For horizontal equilibrium in mid-air, the magnetic force must balance the gravitational force: I L B = m g. Therefore, I = (m g) / (L B) = (0.050 kg × 9.8 m s⁻²) / (0.5 m × 0.4 T) = 0.49 / 0.2 = 2.45 A.

medium

4. Consider the following two statements regarding a charged particle moving in a uniform magnetic field: Statement I: The magnetic force does no work on a moving charged particle. Statement II: The magnetic force can change the direction of motion but cannot change the kinetic energy of the particle.

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

The magnetic force F = q(v × B) is always perpendicular to velocity v. Therefore, power P = F · v = 0, which means no work is done on the particle (Statement I is correct). By the work-energy theorem, since W = ΔK = 0, the kinetic energy remains constant while the direction changes (Statement II is correct).

easy

5. A square wire frame of side L carries a current I. The frame is positioned in a uniform magnetic field B directed along the +z-axis. The vertices of the square are at (0,0,0), (L,0,0), (L,L,0), and (0,L,0) with current circulating counterclockwise in the xy-plane. The net magnetic force on the complete frame is:

  1. Zero
  2. 4 I L B
  3. I L B
  4. 2 I L B
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Answer: (A) Zero

For any closed current-carrying planar loop placed in a uniform magnetic field, the net magnetic force is given by F = I (∮ dl) × B. Since the vector sum of displacement vectors around any closed path is zero (∮ dl = 0), the net magnetic force on the closed square loop is strictly zero.

hard

6. A particle of mass m and charge +q is projected from the origin with velocity v = v₀ î into a region of space where a uniform magnetic field B = B₀ ĵ is present between x = 0 and x = d. If d = (m v₀) / (2 q B₀), the angle with the x-axis at which the particle emerges from the magnetic field is:

  1. 60°
  2. 45°
  3. 0°
  4. 30°
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Answer: (D) 30°

The particle moves in a circular path in the xy-plane with radius R = (m v₀) / (q B₀). The width of the field region is d = R / 2. If the particle deviates by an angle θ upon emerging, then from geometry sin θ = d / R = (R / 2) / R = 1/2. Therefore, θ = 30°.

easy

7. A 0.2 m straight conductor carrying 9 A lies in a uniform 1 T magnetic field at 30° to the field. The force on it is:

  1. 0.9 N
  2. 45 N
  3. 1.8 N
  4. 3.6 N
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Answer: (A) 0.9 N

F = BIL sin θ = 1 × 9 × 0.2 × sin 30° = 0.9 N.

easy

8. A charged particle of mass m and charge q enters a region of uniform magnetic field B with velocity v directed at an angle θ (0° < θ < 90°) to the field lines. If the particle executes a helical path, the ratio of the pitch of the helix to its radius of curvature is:

  1. 2π tan θ
  2. 2π sin θ
  3. 2π cot θ
  4. 2π cos θ
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Answer: (C) 2π cot θ

The pitch of the helix is given by p = v_parallel × T = (v cos θ) × (2πm / (qB)). The radius of the circular cross-section is r = (m v_perp) / (qB) = (m v sin θ) / (qB). Dividing pitch by radius gives p / r = [2πm v cos θ / (qB)] / [m v sin θ / (qB)] = 2π cot θ.

easy

9. Given below are two statements: Statement I: A charged particle entering a region of uniform magnetic field at an acute angle to the field lines follows a helical path with a constant pitch. Statement II: The magnetic force changes both the component of velocity parallel to the magnetic field and the component perpendicular to the magnetic field. In the light of the above statements, choose the correct answer:

  1. Statement I is true but Statement II is false
  2. Statement I is false but Statement II is true
  3. Both Statement I and Statement II are true
  4. Both Statement I and Statement II are false
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Answer: (A) Statement I is true but Statement II is false

Statement I is true because when a charged particle moves at an acute angle to a uniform magnetic field, its parallel velocity remains constant while the perpendicular component undergoes uniform circular motion, resulting in a helical path with constant pitch. Statement II is false because the magnetic force F = q(v × B) is always perpendicular to B, so it has no component along B and cannot change the parallel velocity component.

easy

10. A 0.1 m straight conductor carrying 10 A lies in a uniform 1 T magnetic field at 90° to the field. The force on it is:

  1. 100 N
  2. 1 N
  3. 0.5 N
  4. 2 N
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Answer: (B) 1 N

F = BIL sin θ = 1 × 10 × 0.1 × sin 90° = 1 N.

easy

11. A horizontal wire of mass per unit length 0.04 kg m⁻¹ carries a steady current of 2 A along the positive x-direction. What is the minimum magnitude and direction of the uniform magnetic field required to support the weight of the wire against gravity? (Take g = 10 m s⁻²)

  1. 0.2 T along the positive y-direction
  2. 0.2 T along the positive z-direction
  3. 0.4 T along the positive y-direction
  4. 0.4 T along the negative z-direction
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Answer: (A) 0.2 T along the positive y-direction

To balance gravity, the magnetic force per unit length must act vertically upwards along the positive z-axis: (F/L) = I(î × B) = mg/L k̂. With current along î, we need î × B = k̂, so B must point along the positive y-direction (ĵ). The magnitude is B = (m/L)g / I = (0.04 × 10) / 2 = 0.2 T.

easy

12. A beam of protons enters a uniform magnetic field with speed v at an angle of 90° to the field lines and traverses a circular path of radius R. If the speed of the protons is doubled and the magnetic field strength is tripled, what is the new radius of the circular path?

  1. (2/3) R
  2. (3/2) R
  3. (4/3) R
  4. (1/6) R
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Answer: (A) (2/3) R

The radius of the circular path of a charged particle in a magnetic field is given by R = mv / (qB). When speed becomes 2v and magnetic field becomes 3B, the new radius is R' = m(2v) / [q(3B)] = (2/3) (mv / qB) = (2/3) R.

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