Moving Charges and Magnetism NEET PYQs (2005–2026) with Solutions

10 solved Moving Charges and Magnetism PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumMagnetic field at centre of circular arc

   A current $I_0$ flows through a metallic circular loop of radius $r$ as shown in the figure. Resistance of the segment $ABC$ is half that of $ADC$. Magnitude of magnetic field at the centre $O$ of the loop is :

   1. AOption A: $\frac{\mu_0 I_0}{12r}$
   2. BOption B: $\frac{\mu_0 I_0}{4r}$
   3. COption C: $\frac{\mu_0 I_0}{2r}$
   4. DOption D: $\frac{\mu_0 I_0}{2\pi r}$

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   Correct answer: (A) $\frac{\mu_0 I_0}{12r}$

   Explanation

   Arcs $ABC$ and $ADC$ are in parallel between $A$ and $C$, so the current divides inversely to resistance: $I_{ABC} = \frac{2I_0}{3}$ and $I_{ADC} = \frac{I_0}{3}$. $A$ and $C$ are diametrically opposite, so each arc is a semicircle. Field of a semicircular arc at its centre: $B = \frac{\mu_0 I}{4r}$. The two currents go round $O$ in opposite senses, so their fields oppose. The straight leads lie along the line through $O$, so they give no field at $O$. $B = \frac{\mu_0}{4r}\left(\frac{2I_0}{3} - \frac{I_0}{3}\right) = \frac{\mu_0 I_0}{12r}$, option (A).

2. Question 2 (NEET 2026)

   NEET 2026MediumForce 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.]

   1. AOption A: $\frac{\mu_0 I^2}{2\pi\lambda g}$
   2. BOption B: $\frac{\mu_0 I^2}{\pi\lambda g}$
   3. COption C: $\frac{2\mu_0 I^2}{\pi\lambda g}$
   4. DOption D: $\frac{4\mu_0 I^2}{\pi\lambda g}$

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   Correct answer: (B) $\frac{\mu_0 I^2}{\pi\lambda g}$

   Explanation

   Parallel currents in the same direction attract, so wire $A$ is pulled upward towards $B$. Force per unit length: $f = \frac{\mu_0 I_A I_B}{2\pi h} = \frac{\mu_0 I(2I)}{2\pi h} = \frac{\mu_0 I^2}{\pi h}$. Wire $A$ does not rise if $f \le \lambda g$: $\frac{\mu_0 I^2}{\pi h} \le \lambda g \Rightarrow h \ge \frac{\mu_0 I^2}{\pi\lambda g}$. Minimum $h = \frac{\mu_0 I^2}{\pi\lambda g}$, option (B).

3. Question 3 (NEET 2026)

   NEET 2026MediumMagnetic field due to current carrying circular loop and magnetic dipole moment

   A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take μ₀ = 4π × 10⁻⁷ T m/A)

   1. AOption A: 2 A, 10 A m²
   2. BOption B: 2.5 A, 20 A m²
   3. COption C: 2 A, 4 A m²
   4. DOption D: 2.5 A, 2 A m²

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   Correct answer: (D) 2.5 A, 2 A m²

   Explanation

   Magnetic field at centre of coil: B=μ₀Ni/2R. Therefore i=(2RB)/(μ₀N)=2.5 A. Magnetic moment M=NiA=100×2.5×π×(0.05)²≈2 A m². Hence option (D) is correct.

4. Question 4 (NEET 2026)

   NEET 2026EasyMoving Coil Galvanometer

   A galvanometer of resistance 100 Ω gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0–10 A. The shunt required is:

   1. AOption A: 0.01 Ω
   2. BOption B: 0.10 Ω
   3. COption C: 1.0 Ω
   4. DOption D: 0.001 Ω

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   Correct answer: (A) 0.01 Ω

   Explanation

   Galvanometer current Ig=1 mA=0.001 A, resistance Rg=100 Ω. Shunt current Is≈10 A. Since IgRg=IsRs, Rs=(0.001×100)/10=0.01 Ω. Hence option (A) is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumMagnetic field due to current carrying conductor (Ampere's law)

   The figure given below shows a long straight solid wire of circular cross-section of radius 'a' carrying steady current I. The current I is uniformly distributed across its cross-section. The plot which correctly represents the variation of magnetic field (B) with distance (r) from the axis of the conductor in the region is:

   1. AOption A: Graph 1
   2. BOption B: Graph 2
   3. COption C: Graph 3
   4. DOption D: Graph 4

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   Correct answer: (A) Graph 1

   Explanation

   Inside the conductor (r<a), magnetic field B=(μ₀Ir)/(2πa²), therefore B∝r. Outside the conductor (r>a), B=μ₀I/(2πr), therefore B∝1/r. Hence magnetic field increases linearly inside and decreases inversely outside, matching graph (1).

6. Question 6 (NEET 2025)

   NEET 2025MediumMagnetic Force

   An electron (mass $9\times10^{-31}\,\text{kg}$ and charge $1.6\times10^{-19}\,\text{C}$) moving with speed $\dfrac{c}{100}$ ($c$ = speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9\times10^{-4}\,\text{T}$ perpendicular to its direction of motion. We wish to apply a uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then ($c = 3\times10^8\,\text{m s}^{-1}$)

   1. AOption A: $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^{4}\,\text{V m}^{-1}$
   2. BOption B: $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^{2}\,\text{V m}^{-1}$
   3. COption C: $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27\times10^{2}\,\text{V m}^{-1}$
   4. DOption D: $\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27\times10^{4}\,\text{V m}^{-1}$

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   Correct answer: (B) $\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27\times10^{2}\,\text{V m}^{-1}$

   Explanation

   For the electron to move undeflected, the electric force must balance the magnetic force: $$qE = qvB$$ Therefore, $$E = vB$$ Given: $$v = \frac{c}{100} = \frac{3\times10^8}{100} = 3\times10^6\,\text{m s}^{-1}$$ and $$B = 9\times10^{-4}\,\text{T}$$ Hence, $$E = (3\times10^6)(9\times10^{-4})$$ $$E = 27\times10^2\,\text{V m}^{-1}$$ The electric field must be perpendicular to the magnetic field so that electric and magnetic forces act in opposite directions. Hence, the correct option is (2).

7. Question 7 (NEET 2025)

   NEET 2025EasyMagnetic Dipole & Torque

   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 :

   1. AOption A: 1:4
   2. BOption B: 1:2
   3. COption C: 2:1
   4. DOption D: 4:1

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   Correct answer: (A) 1:4

   Explanation

   Magnetic moment of a circular coil is M = IA = Iπr². Since current is same in both coils, magnetic moment is proportional to r². Therefore ratio = 1² : 2² = 1 : 4.

8. Question 8 (NEET 2024)

   NEET 2024MediumMagnetic dipole; Oscillations of a magnetic needle in uniform magnetic field

   In a uniform magnetic field of $0.049\,\text{T}$, a magnetic needle performs $20$ complete oscillations in $5\,\text{s}$ as shown. The moment of inertia of the needle is $9.8\times10^{-6}\,\text{kg m}^2$. If the magnetic moment of the needle is $x\times10^{-5}\,\text{A m}^2$, then the value of $x$ is:

   1. AOption A: $50\pi^2$
   2. BOption B: $1280\pi^2$
   3. COption C: $5\pi^2$
   4. DOption D: $128\pi^2$

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   Correct answer: (B) $1280\pi^2$

   Explanation

   Time period: $$T=\frac{5}{20}=0.25\,s$$ For oscillations of a magnetic needle: Rearranging: $$M=\frac{4\pi^2 I}{B T^2}$$ Substituting: $$M=\frac{4\pi^2(9.8\times10^{-6})}{(0.049)(0.25)^2}$$ $$M=1280\pi^2\times10^{-5}\,\text{A m}^2$$ Hence: $$x=1280\pi^2$$ Therefore, the correct answer is: **(2) $1280\pi^2$**

9. Question 9 (NEET 2024)

   NEET 2024EasyBiot-Savart Law

   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):

   1. AOption A: 4.4 mT
   2. BOption B: 44 T
   3. COption C: 44 mT
   4. DOption D: 4.4 T

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   Correct answer: (A) 4.4 mT

   Explanation

   Magnetic field at the centre of a circular coil is: $$B = \frac{\mu_0 NI}{2R}$$ Given: $$\mu_0 = 4\pi \times 10^{-7}$$ $$N = 100$$ $$I = 7\,A$$ $$R = 10\,cm = 0.1\,m$$ Substituting: $$B = \frac{4\pi \times 10^{-7} \times 100 \times 7}{2 \times 0.1}$$ $$B \approx 4.4 \times 10^{-3}\,T$$ $$B = 4.4\,mT$$ Hence, option (A) is correct.

10. Question 10 (NEET 2023)

    NEET 2023MediumForce on Current-Carrying Conductors

    A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})\,\text{T}$. The magnitude of the magnetic force acting on the wire is:

    1. AOption A: $3IL$
    2. BOption B: $\sqrt{5}\,IL$
    3. COption C: $5IL$
    4. DOption D: $\sqrt{3}\,IL$

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    Correct answer: (C) $5IL$

    Explanation

    Magnetic force on a current carrying wire is given by: $$\vec{F} = I(\vec{L} \times \vec{B})$$ Since the wire is along the positive $x$-axis:$$\vec{L} = L\hat{i}$$Given magnetic field:$$\vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k}$$Cross product:$$\vec{L} \times \vec{B} = L\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k})$$$$= L(0 + 3\hat{k} + 4\hat{j})$$Magnitude:$$|\vec{L} \times \vec{B}| = L\sqrt{3^2 + 4^2} = 5L$$Therefore,$$F = I(5L) = 5IL$$ Hence, option (C) is correct.