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Current loop/bar magnet as magnetic dipoles and magnetic fields: 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 planar circular loop of area A and carrying current I is placed in a uniform external magnetic field B. Which of the following statements is correct regarding the potential energy U of the loop?

  1. U is minimum when the magnetic moment M is antiparallel to B
  2. U is minimum when the magnetic moment M is parallel to B
  3. U is zero when the magnetic moment M is antiparallel to B
  4. U is maximum when the magnetic moment M is perpendicular to B
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Answer: (B) U is minimum when the magnetic moment M is parallel to B

The potential energy of a magnetic dipole in a magnetic field is given by U = -M · B = -M B cos θ. When M is parallel to B (θ = 0°), U = -M B, which corresponds to the minimum potential energy and stable equilibrium.

medium

2. A circular coil of radius R and N turns carries a current I. The ratio of the magnitude of its magnetic dipole moment to the magnitude of the magnetic field at its centre is proportional to:

  1. R³
  2. R²
  3. R
  4. R⁴
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Answer: (A) R³

The magnetic dipole moment is M = N I (π R²). The magnetic field at the centre of the coil is B = μ₀ N I / (2R). The ratio M / B = [N I π R²] / [μ₀ N I / (2R)] = (2π / μ₀) R³. Thus, the ratio is proportional to R³.

hard

3. Two identical short magnetic dipoles of magnetic moments M each are placed at the origin such that their magnetic moments are directed along the +x and +y axes respectively. The magnitude of the resultant magnetic field at a distant point (d, 0, 0) on the x-axis is: (Take constant μ₀ / 4π = k)

  1. k M / d³
  2. 2 k M / d³
  3. 3 k M / d³
  4. √5 k M / d³
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Answer: (D) √5 k M / d³

For the dipole along the +x axis, the point (d, 0, 0) lies on its axial line, producing a field B₁ = 2 k M / d³ along the +x direction. For the dipole along the +y axis, the point lies on its equatorial line, producing a field B₂ = k M / d³ antiparallel to the +y axis. Since B₁ and B₂ are perpendicular, the resultant field magnitude is B = √(B₁² + B₂²) = (k M / d³) √(2² + 1²) = √5 k M / d³.

medium

4. Gauss's law for magnetism states that the net magnetic flux through any closed surface is:

  1. Equal to μ₀ times enclosed current
  2. Equal to enclosed charge/ε₀
  3. Infinite
  4. Zero
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Answer: (D) Zero

Magnetic field lines form closed loops and isolated magnetic poles (monopoles) do not exist, so the net magnetic flux through any closed surface is zero.

NCERT Class 12

easy

5. Consider the following two statements regarding a current-carrying loop and a magnetic dipole: Statement I: A planar current-carrying loop of any arbitrary shape behaves as a magnetic dipole with dipole moment vector M = I A, where the direction of area vector A is given by the right-hand thumb rule. Statement II: Unlike an electric dipole, the magnetic field lines of a magnetic dipole form continuous closed loops without start or end. 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: (A) Both Statement I and Statement II are correct

Statement I is correct because any planar current loop behaves as a magnetic dipole of moment M = I A, with direction normal to the plane determined by the right-hand rule. Statement II is also correct because isolated magnetic poles (monopoles) do not exist, so magnetic field lines are continuous closed loops.

hard

6. A non-conducting disc of mass m and total positive charge q distributed uniformly over its surface is spun about its central symmetry axis with angular velocity ω. The ratio of its magnetic dipole moment M to its angular momentum L is:

  1. q / (2m)
  2. q / m
  3. 2q / m
  4. q / (4m)
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Answer: (A) q / (2m)

For any body where charge and mass are distributed uniformly in the same geometric manner, the gyromagnetic ratio is M / L = q / (2m). Specifically for a uniform disc, M = (1/4) q ω R² and L = (1/2) m ω R², yielding M / L = q / (2m).

medium

7. Given below are two statements: Statement I: The net torque acting on a planar current loop placed in a uniform magnetic field depends on the shape of the loop as long as the area remains constant. Statement II: The potential energy of a magnetic dipole of dipole moment M placed in a uniform magnetic field B is minimum when M and B are parallel to each other. In the light of the above statements, choose the most appropriate answer from the options given below:

  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: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because torque τ = M × B = I A × B, which depends only on the current and the planar area A, not the specific geometric shape of the loop. Statement II is correct because U = -M · B = -M B cos θ, which attains its minimum value (-M B) when θ = 0° (parallel orientation).

medium

8. A point P lies at a distance r from the centre of a short magnetic dipole of moment M, at an angle θ with the dipole axis. The angle α that the total magnetic field vector at P makes with the position vector r is given by:

  1. tan α = tan θ
  2. tan α = (1/2) tan θ
  3. tan α = 2 tan θ
  4. tan α = cot θ
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Answer: (B) tan α = (1/2) tan θ

The magnetic field due to a dipole has radial component B_r = (μ₀ / 4π) (2M cos θ / r³) and transverse component B_θ = (μ₀ / 4π) (M sin θ / r³). The angle α made by the resultant field with the position vector r is given by tan α = B_θ / B_r = (M sin θ) / (2M cos θ) = (1/2) tan θ.

easy

9. An electron of charge e and mass m moves in a circular orbit of radius r with speed v. The ratio of its orbital magnetic dipole moment to its orbital angular momentum is:

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

The orbital magnetic dipole moment is M = I A = (e v / 2π r) (π r²) = e v r / 2. The orbital angular momentum is L = m v r. The gyromagnetic ratio is M / L = (e v r / 2) / (m v r) = e / (2m), which is a fundamental constant for orbital motion.

easy

10. A current-carrying solenoid behaves like a:

  1. Point charge
  2. Capacitor
  3. Bar magnet
  4. Electric dipole
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Answer: (C) Bar magnet

The field of a finite solenoid resembles that of a bar magnet; one end acts as a north pole and the other as a south pole.

NCERT Class 12

medium

11. Two identical magnetic dipoles of magnetic moment M are placed at right angles to each other with their south poles touching at a common point O. The magnitude of the resultant magnetic moment of the system is:

  1. M
  2. 2 M
  3. √2 M
  4. M / √2
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Answer: (C) √2 M

Magnetic dipole moment is a vector quantity pointing from the south pole to the north pole. Here, two perpendicular vectors of magnitude M originate from point O. The resultant magnetic moment is M_net = √(M² + M² + 2M² cos 90°) = √(2M²) = √2 M.

medium

12. A short bar magnet with magnetic moment M is oriented along the x-axis. The magnetic field produced by this magnet at a point (0, y, 0) on the y-axis has a magnitude B. If the magnet is rotated by 90° in the xy-plane so that it points along the y-axis, the magnitude of the magnetic field at the same point (0, y, 0) becomes:

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

Initially, the point (0, y, 0) lies on the equatorial line of the dipole, so B_initial = (μ₀/4π)(M/y³) = B. After rotating the magnet by 90° to lie along the y-axis, the point (0, y, 0) lies on its axial line. The field at an axial point at the same distance is B_final = (μ₀/4π)(2M/y³) = 2B.

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