Magnetic Effects of Current and Magnetism: JEE Main 2026 PYQs

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Answer: 171

Key idea: Lorentz Force and Cross Product of Vectors

1. The magnetic force on a charged particle is $\vec{F} = q(\vec{v} \times \vec{B})$ where $q = 1 \times 10^{-6}$ C
2. Calculate cross product: $\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$
3. $\vec{v} \times \vec{B} = \hat{i}(10-9) - \hat{j}(-5-6) + \hat{k}(3+4) = \hat{i} + 11\hat{j} + 7\hat{k}$
4. Force vector: $\vec{F} = 10^{-6}(\hat{i} + 11\hat{j} + 7\hat{k})$ N
5. Magnitude: $|\vec{F}| = 10^{-6}\sqrt{1^2 + 11^2 + 7^2} = 10^{-6}\sqrt{171}$ N
6. Comparing with $\sqrt{\alpha} \times 10^{-6}$ N, we get $\alpha = 171$

Answer: Option (1)

Key idea: Lorentz Force on a Charged Particle

1. Write the Lorentz force equation: $\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})$
2. Calculate electric force: $\vec{F}_E = 10^{-9} \times 0.4\hat{j} = 4 \times 10^{-10}\hat{j}$ N
3. Let velocity be $\vec{v} = v_x\hat{i} + v_y\hat{j}$. Calculate cross product: $\vec{v} \times \vec{B} = (v_x\hat{i} + v_y\hat{j}) \times (4 \times 10^{-3}\hat{k}) = 4 \times 10^{-3}(v_y\hat{i} - v_x\hat{j})$
4. Magnetic force: $\vec{F}_B = 10^{-9} \times 4 \times 10^{-3}(v_y\hat{i} - v_x\hat{j}) = 4 \times 10^{-12}(v_y\hat{i} - v_x\hat{j})$
5. Total force: $\vec{F} = 4v_y \times 10^{-12}\hat{i} + (4 \times 10^{-10} - 4v_x \times 10^{-12})\hat{j}$
6. Compare with given force $(4\hat{i} + 2\hat{j}) \times 10^{-10}$ N: From $\hat{i}$ component: $v_y = 100$ m/s; From $\hat{j}$ component: $v_x = 50$ m/s
7. Therefore, $\vec{v} = 50\hat{i} + 100\hat{j}$ m/s

Answer: Option (2)

Key idea: Magnetic moment of a flat coil with non-uniform radii; requires averaging over continuous radius distribution

1. Given: $N = 200$ turns, $r_1 = 3$ cm $= 0.03$ m, $r_2 = 6$ cm $= 0.06$ m, $I = 20$ mA $= 0.02$ A
2. For uniformly distributed radii from $r_1$ to $r_2$, find average $r^2$: $\langle r^2 \rangle = \frac{r_2^3 - r_1^3}{3(r_2 - r_1)}$
3. Calculate: $\langle r^2 \rangle = \frac{(0.06)^3 - (0.03)^3}{3(0.06 - 0.03)} = \frac{0.000216 - 0.000027}{0.09} = 0.0021$ m$^2$
4. Average area: $A_{avg} = \pi \langle r^2 \rangle = \pi \times 0.0021 = 0.006597$ m$^2$
5. Magnetic moment: $\mu = NIA_{avg} = 200 \times 0.02 \times 0.006597 = 0.0264$ A·m$^2$
6. Express as $\mu = 2.64 \times 10^{-2}$ A·m$^2$, so $\alpha = 2.64$

Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.

Key idea: Relationship between magnetic field B, magnetic intensity H, and permeability in magnetic materials

Answer: 12

Key idea: Magnetic field due to bar magnet at axial and equatorial points; Vector addition of perpendicular fields

1. Point P is at distance $r = 5$ cm $= 0.05$ m from each magnet
2. For left magnet (horizontal): P is on axial line, so $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{r^3} = 10^{-7} \cdot \frac{2 \times 3\sqrt{5}}{(0.05)^3} = 4.8\sqrt{5} \times 10^{-3}$ T (horizontal)
3. For right magnet (vertical): P is on equatorial line, so $B_2 = \frac{\mu_0}{4\pi} \cdot \frac{M}{r^3} = 10^{-7} \cdot \frac{3\sqrt{5}}{(0.05)^3} = 2.4\sqrt{5} \times 10^{-3}$ T (vertical)
4. Since $B_1$ and $B_2$ are perpendicular: $B_{net} = \sqrt{B_1^2 + B_2^2}$
5. $B_{net} = 10^{-3}\sqrt{(4.8\sqrt{5})^2 + (2.4\sqrt{5})^2} = 10^{-3}\sqrt{115.2 + 28.8} = 10^{-3}\sqrt{144}$
6. $B_{net} = 12 \times 10^{-3}$ T, therefore $a = 12$

Answer: 314

Key idea: Torque on a current-carrying coil in a magnetic field

1. Write the torque formula: $\tau = NIAB\sin\theta$ where $N$ is turns, $I$ is current, $A$ is area, $B$ is magnetic field, and $\theta$ is the angle
2. Calculate the area of circular coil: $A = \pi r^2 = 3.14 \times (0.02)^2 = 0.001256$ m²
3. Note that $\sin(30°) = 0.5$
4. Substitute values: $\tau = 125 \times 1 \times 0.001256 \times 0.4 \times 0.5$
5. Calculate: $\tau = 125 \times 0.0002512 = 0.0314$ N-m
6. Express in required form: $\tau = 314 \times 10^{-4}$ N-m, so $a = 314$

Answer: 12

Key idea: Magnetic force on a charged particle is perpendicular to both velocity and magnetic field

1. Given: $\vec{B} = (3\hat{i} + 2\hat{j})$ T and $\vec{a} = (4\hat{i} - \frac{x}{2}\hat{j})$ m/s²
2. Magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to $\vec{B}$
3. Since $\vec{F} = m\vec{a}$, the acceleration $\vec{a}$ must also be perpendicular to $\vec{B}$
4. Apply perpendicularity condition: $\vec{a} \cdot \vec{B} = 0$
5. Calculate: $(4\hat{i} - \frac{x}{2}\hat{j}) \cdot (3\hat{i} + 2\hat{j}) = 12 - x = 0$
6. Solve: $x = 12$

Answer: Option (1)

Key idea: Work-Energy Theorem with Electromagnetic Forces

1. Apply work-energy theorem: $\Delta KE = W_{total}$
2. Note that magnetic force does no work since $\vec{F}_B \perp \vec{v}$ always
3. Only electric force contributes: $W = \int_0^L q\vec{E} \cdot d\vec{r}$
4. Since $\vec{E} = E_0 e^{-\lambda x}\hat{x}$, we get $W = qE_0 \int_0^L e^{-\lambda x} dx$
5. Evaluate integral: $W = qE_0 \left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^L$
6. Simplify: $\Delta KE = \frac{qE_0}{\lambda}\left[1 - e^{-\lambda L}\right]$

Answer: Option (1)

Key idea: Magnetic energy density and magnetic field at the center of a circular current loop

1. Find magnetic field at center of loop: $B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 1.256 \times 10^{-5}$ T
2. Calculate volume of cube: $V = a^3 = (10^{-3})^3 = 10^{-9}$ m³
3. Find magnetic energy density: $u = \frac{B^2}{2\mu_0} = \frac{(1.256 \times 10^{-5})^2}{2 \times 4\pi \times 10^{-7}} = 6.28 \times 10^{-5}$ J/m³
4. Calculate total energy: $U = u \times V = 6.28 \times 10^{-5} \times 10^{-9} = 6.28 \times 10^{-14}$ J
5. Compare with $a \times 10^{-14}$ J to get $a = 6.28$

Answer: Option (300)

Key idea: Magnetic field due to parallel current-carrying wires; superposition principle

1. Distance from each wire to midpoint: $r = \frac{8}{2} = 4$ cm $= 0.04$ m
2. Magnetic field due to one wire: $B_1 = \frac{\mu_0}{4\pi} \times \frac{2I}{r} = 10^{-7} \times \frac{2 \times 30}{0.04}$
3. Calculate: $B_1 = 10^{-7} \times 1500 = 1.5 \times 10^{-4}$ T $= 150$ μT
4. Since currents are opposite, both fields at midpoint point in the same direction (use right-hand rule)
5. Total magnetic field: $B_{net} = B_1 + B_2 = 150 + 150 = 300$ μT

Answer: 50

Key idea: Galvanometer with shunt and series resistance; current division and Ohm's law

1. Let $G$ be the galvanometer resistance and $I_g$ be the full-scale deflection current.
2. **Shunt case:** When shunted with $2\,\Omega$, total current is $0.5\,\text{A}$. Using current division: $I_g = \frac{1}{2+G}$
3. **Series case:** With $470\,\Omega$ in series and $10\,\text{V}$ applied: $I_g = \frac{10}{G+470}$
4. Equate both expressions for $I_g$: $\frac{1}{2+G} = \frac{10}{G+470}$
5. Cross-multiply: $G + 470 = 10(2+G) = 20 + 10G$
6. Solve: $450 = 9G \Rightarrow G = 50\,\Omega$

Answer: Option (1)

Key idea: Torque on a current-carrying loop in a magnetic field

1. Calculate the area of the loop: $A = \pi r^2 = 3.14 \times (0.02)^2 = 1.256 \times 10^{-3}$ m²
2. Find the magnetic moment vector: $\vec{M} = IA\hat{n} = 100\sqrt{2} \times 1.256 \times 10^{-3} \times \frac{\hat{i}+\hat{k}}{\sqrt{2}} = 0.1256(\hat{i}+\hat{k})$ A·m²
3. Express the magnetic field: $\vec{B} = 4 \times 10^{-3}(3\hat{i}+2\hat{k}) = (0.012\hat{i}+0.008\hat{k})$ T
4. Calculate torque using $\vec{\tau} = \vec{M} \times \vec{B}$
5. Evaluate the cross product: $\vec{\tau} = 0.1256 \times 0.008(\hat{i} \times \hat{k}) + 0.1256 \times 0.012(\hat{k} \times \hat{i})$
6. Simplify: $\vec{\tau} = -0.001005\hat{j} + 0.001507\hat{j} = 5.02 \times 10^{-4}\hat{j}$ Wb·A

Answer: 2

Key idea: Helical motion of charged particle in magnetic field

1. Identify velocity components: $v_\perp = 3 \times 10^{-2}$ m/s (perpendicular to $\vec{B}$) and $v_\parallel = 2 \times 10^{-2}$ m/s (parallel to $\vec{B}$)
2. The particle undergoes helical motion: circular motion in xy-plane + linear motion along z-axis
3. Calculate time period: $T = \frac{2\pi m}{qB} = \frac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1} = 20$ s
4. Time for 5 revolutions: $t = 5T = 5 \times 20 = 100$ s
5. Distance along $\hat{k}$ (pitch): $a = v_\parallel \times t = 2 \times 10^{-2} \times 100 = 2$ m