NEET Moving Charges and Magnetism PYQs & Practice Questions

10 free NEET Physics questions on Moving Charges and Magnetism, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Velocity Selector · Magnetic Moment · Biot-Savart Law · Ampere's Circuital Law · Motion of Charged Particle in Magnetic Field · Magnetic Field on Axis of Circular Loop · Force Between Parallel Current-Carrying Conductors · Moving Coil Galvanometer · Helical Motion in a Magnetic Field · Biot-Savart Law and Field Direction · Charged Particle Moving Along a Solenoid's Axis · Energy Gain in a Cyclotron

1. Q1NEET 2026Force Between Parallel Current-Carrying Wires

   Two infinitely long parallel conducting wires A and B carry currents I and 2I, respectively, in the same direction. The wire A has uniform mass per unit length lambda and lies on an insulated floor. The wire B is kept fixed at a height h above the floor. The minimum magnitude of h so that the wire A does not rise from the floor is: [g is the acceleration due to gravity and mu_0 is the permeability of free space.]

   • A. mu_0*I^2/(pi*lambda*g)
   • B. 2*mu_0*I^2/(pi*lambda*g)
   • C. 4*mu_0*I^2/(pi*lambda*g)
   • D. mu_0*I^2/(2*pi*lambda*g)
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   Answer: —

   Since both wires carry current in the same direction, they attract each other, so wire B (above) pulls wire A upward. Force per unit length = mu_0 * I * (2I) / (2*pi*h) = mu_0*I^2/(pi*h). For A to just not rise, this must equal the weight per unit length lambda*g: mu_0*I^2/(pi*h) = lambda*g, giving h = mu_0*I^2/(pi*lambda*g).

2. Q2NEET 2026Magnetic Field at Centre of a Current-Carrying Circular Arc

   A current I₀ flows through a metallic circular loop of radius r. The current enters the loop at point A (top) and leaves at point C (bottom, diametrically opposite A), splitting into two semicircular paths: A-B-C (through B on the left) and A-D-C (through D on the right). The resistance of the segment ABC is half that of ADC. The magnitude of the magnetic field at the centre O of the loop is:

   • A. μ₀I₀/4r
   • B. μ₀I₀/2r
   • C. μ₀I₀/2πr
   • D. μ₀I₀/12r
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   Answer: C — μ₀I₀/2πr

   Since R_ABC = R_ADC/2, current divides inversely to resistance: I_ABC = 2I_ADC, and since I_ABC + I_ADC = I₀, I_ADC = I₀/3 and I_ABC = 2I₀/3. Both ABC and ADC are semicircular arcs (A and C are diametrically opposite), and a semicircular arc carrying current I produces field μ₀I/(4r) at the centre. The two currents circulate in opposite senses (one clockwise, one counter-clockwise as seen from the centre), so their fields oppose: B_net = μ₀(2I₀/3)/(4r) - μ₀(I₀/3)/(4r) = μ₀I₀/(6r) - μ₀I₀/(12r) = μ₀I₀/(12r).

3. Q3NEET 2025Velocity Selector

   An electron (mass 9×10⁻³¹ kg and charge 1.6×10⁻¹⁹C) moving with speed c/100 (c = speed of light) is injected into a magnetic field B of magnitude 9×10⁻⁴ T perpendicular to its direction of motion. We wish to apply a uniform electric field E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3×10⁸ ms⁻¹)

   • A. E is perpendicular to B and its magnitude is 27×10⁴ Vm⁻¹
   • B. E is perpendicular to B and its magnitude is 27×10² Vm⁻¹
   • C. E is parallel to B and its magnitude is 27×10² Vm⁻¹
   • D. E is parallel to B and its magnitude is 27×10⁴ Vm⁻¹
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   Answer: A — E is perpendicular to B and its magnitude is 27×10⁴ Vm⁻¹

   For no deflection, electric force must balance magnetic force: qE = qvB, so E = vB = (c/100)×B = (3×10⁸/100)×9×10⁻⁴ = 27×10² Vm⁻¹, with E perpendicular to B (and to v).

4. Q4NEET 2025Magnetic Moment

   A 2 amp current is flowing through two different small circular copper coils having radii ratio 1:2. The ratio of their respective magnetic moments will be

   • A. 1:4
   • B. 1:2
   • C. 2:1
   • D. 4:1
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   Answer: —

   Magnetic moment M = I·π·r². Since current is the same, M1/M2 = (r1/r2)² = (1/2)² = 1/4.

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5. Q5NEET 2024Magnetic Field at Centre of Current-Carrying Coil

   A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π×10⁻⁷ SI units):

   • A. 4.4 mT
   • B. 44 T
   • C. 44 mT
   • D. 4.4 T
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   Answer: —

   B = μ0NI/2R = (4π×10⁻⁷ × 100 × 7)/(2 × 0.1) = 4.4 mT.

6. Q6NEET 2023Magnetic Field Due to Semi-Circular Wire

   A very long conducting wire is bent in a semi-circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by: [Figure shows two long straight wire segments with current i connected by a semicircular arc of radius R, point P at the center of the semicircle]

   • A. μ0i/4R pointed away from the page
   • B. (μ0i/4R)[1-2/π] pointed away from page
   • C. (μ0i/4R)[1-2/π] pointed into the page
   • D. μ0i/4R pointed into the page
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   Answer: A — μ0i/4R pointed away from the page

   The total field at P is the sum of contributions from the two straight segments and the semicircular arc. Combining these contributions gives Bp = (μ0i/4R)[1-2/π], pointed away from the page.

7. Q7NEET 2023Magnetic Force on Current-Carrying Wire

   A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field B = (2î+3ĵ-4k̂) T. The magnitude of the magnetic force acting on the wire is:

   • A. √5 IL
   • B. 5 IL
   • C. √3 IL
   • D. 3 IL
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   Answer: A — √5 IL

   F = I(L×B) = I(Lî × (2î+3ĵ-4k̂)) = IL[3k̂+4ĵ] (since î×î=0, î×ĵ=k̂, î×k̂=-ĵ). |F| = IL√(3²+4²) = 5IL.

8. Q8NEET 2022Magnetic Field of a Solenoid

   A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:

   • A. 6.28 x 10^-2 T
   • B. 12.56 x 10^-2 T
   • C. 12.56 x 10^-4 T
   • D. 6.28 x 10^-4 T
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   Answer: A — 6.28 x 10^-2 T

   n = 100 turns/mm = 10^5 turns/m. B = mu0*n*I = 4*pi*10^-7 x 10^5 x 1 = 4*pi*10^-2 = 12.56 x 10^-2 T. (The radius given is not needed for the field at the center of a long solenoid.)

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9. Q9NEET 2022Biot-Savart's Law vs Coulomb's Law

   Given below are two statements: Statement I: Biot-Savart's law gives us the expression for the magnetic field strength of an infinitesimal current element (Idl) of a current carrying conductor only. Statement II: Biot-Savart's law is analogous to Coulomb's inverse square law of charge q, with the former being related to the field produced by a scalar source, Idl while the latter being produced by a vector source, q. In light of the above statements choose the most appropriate answer from the options given below:

   • A. Both Statement I and Statement II are correct
   • B. Both Statement I and Statement II are incorrect
   • C. Statement I is correct and Statement II is incorrect
   • D. Statement I is incorrect and Statement II is correct
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   Answer: B — Both Statement I and Statement II are incorrect

   Statement I correctly describes Biot-Savart's law. Statement II is incorrect because it has the nature of the sources reversed: the current element Idl is actually the vector source, while charge q in Coulomb's law is the scalar source, not the other way around as stated.

10. Q10NEET 2022Magnetic Field Inside and Outside a Current-Carrying Wire

    From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:

    • A. Uniform and remains constant for both the regions.
    • B. A linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.
    • C. A linearly increasing function of distance r upto the boundary of the wire and then decreasing one with 1/r dependence for the outside region.
    • D. A linearly decreasing function of distance upto the boundary of the wire and then a linearly increasing one for the outside region.
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    Answer: B — A linearly increasing function of distance upto the boundary of the wire and then linearly decreasing for the outside region.

    Inside the wire, B is proportional to r (distance from axis) since enclosed current grows with area (r^2) and B = mu0*I_enclosed/(2*pi*r). Outside the wire, B is proportional to 1/r, as for an infinite line current.