Magnetism Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Magnetism with step-by-step solutions covering magnetic field of coils and wires, Lorentz force, magnetic moment, torque on current loops, and charged-particle motion.
Practice the most recent JEE Main 2026 Magnetism questions with clean, step-by-step worked solutions.
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Solution
Each turn contributes $I\pi r^2$ to the magnetic moment. The radii vary uniformly from $r_1$ to $r_2$ across the $N$ turns, so the total moment is
$$M = NI\pi \langle r^2\rangle = NI\pi\cdot\frac{r_1^2 + r_1 r_2 + r_2^2}{3}$$Substituting $N = 200$, $I = 20\times10^{-3}$ A, $r_1 = 0.03$ m, $r_2 = 0.06$ m:
$$r_1^2 + r_1 r_2 + r_2^2 = 0.0009 + 0.0018 + 0.0036 = 0.0063\ \text{m}^2$$$$M = 200 \times 0.02 \times \pi \times \frac{0.0063}{3} = 4\pi \times 0.0021 = 2.64\times10^{-2}\ \text{A.m}^2$$Hence $\alpha = 2.64$.
Answer: B ($2.64$)
Solution
Magnetic field at the centre of the loop:
$$B = \frac{\mu_0 I}{2R} = \frac{4\pi\times10^{-7}\times 2}{2\times 0.10} = 4\pi\times10^{-6} = 1.256\times10^{-5}\ \text{T}$$Magnetic energy density:
$$u = \frac{B^2}{2\mu_0} = \frac{(4\pi\times10^{-6})^2}{2\times 4\pi\times10^{-7}} = \frac{16\pi^2\times10^{-12}}{8\pi\times10^{-7}} = 2\pi\times10^{-5}\ \text{J/m}^3$$Volume of the cube: $V = (10^{-3})^3 = 10^{-9}\ \text{m}^3$. Energy stored:
$$U = uV = 2\pi\times10^{-5}\times10^{-9} = 2\pi\times10^{-14} = 6.28\times10^{-14}\ \text{J}$$Hence $\alpha = 6.28$.
Answer: A ($6.28$)
Solution
The magnetic force is $\vec{F} = q\,\vec{v}\times\vec{B}$. Compute the cross product:
$$\vec{v}\times\vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & 3\\ 2 & 3 & -5\end{vmatrix}$$$$= \hat{i}\big[(-2)(-5) - (3)(3)\big] - \hat{j}\big[(1)(-5) - (3)(2)\big] + \hat{k}\big[(1)(3) - (-2)(2)\big]$$$$= \hat{i}(10 - 9) - \hat{j}(-5 - 6) + \hat{k}(3 + 4) = \hat{i} + 11\hat{j} + 7\hat{k}$$Magnitude:
$$|\vec{v}\times\vec{B}| = \sqrt{1^2 + 11^2 + 7^2} = \sqrt{1 + 121 + 49} = \sqrt{171}$$So $\vec{F} = 10^{-6}\sqrt{171}\ \text{N}$, giving $\sqrt{\alpha} = \sqrt{171}$.
Answer: 171
Solution
The magnetic field inside a medium relates to the magnetic intensity by
$$B = \mu_0 \mu_r H \quad\Rightarrow\quad H = \frac{B}{\mu_0 \mu_r}$$Substituting $B = 1.0$ T, $\mu_r = 400$, $\mu_0 = 4\pi\times10^{-7}$:
$$H = \frac{1.0}{4\pi\times10^{-7}\times 400} = \frac{1}{1600\pi\times10^{-7}} = \frac{10^7}{1600\pi}\ \text{A/m}$$$$H = \frac{10^5}{16\pi} = \frac{1}{16\pi}\times10^5\ \text{A/m}$$Hence $\alpha = \dfrac{1}{16\pi}$.
Answer: B ($\dfrac{1}{16\pi}$)
Solution
The magnetic moment of the coil is
$$m = NIA = NI\pi r^2 = 125 \times 1 \times \pi \times (0.02)^2 = 125\pi\times 4\times10^{-4}$$$$m = 0.05\pi = 0.157\ \text{A.m}^2$$The angle between the coil’s axis (direction of $\vec{m}$) and $\vec{B}$ is $30°$, so the torque is
$$\tau = mB\sin\theta = 0.157 \times 0.4 \times \sin 30° = 0.157 \times 0.4 \times 0.5$$$$\tau = 0.0314\ \text{N.m} = 314\times10^{-4}\ \text{N.m}$$Hence $\alpha = 314$.
Answer: 314
Solution
The Lorentz force is $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$ with $\vec{v} = v_x\hat{i} + v_y\hat{j}$, $\vec{E} = 0.4\hat{j}$, $\vec{B} = 4\times10^{-3}\hat{k}$.
$$\vec{v}\times\vec{B} = (v_x\hat{i} + v_y\hat{j})\times(4\times10^{-3}\hat{k}) = v_y(4\times10^{-3})\hat{i} - v_x(4\times10^{-3})\hat{j}$$Equate components of $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$ with $q = 10^{-9}$:
$$F_x = q\,v_y(4\times10^{-3}) = 4\times10^{-10} \Rightarrow v_y = \frac{4\times10^{-10}}{10^{-9}\times4\times10^{-3}} = 100$$$$F_y = q\big(0.4 - v_x(4\times10^{-3})\big) = 2\times10^{-10}$$$$0.4 - v_x(4\times10^{-3}) = \frac{2\times10^{-10}}{10^{-9}} = 0.2 \Rightarrow v_x(4\times10^{-3}) = 0.2 \Rightarrow v_x = 50$$So $\vec{v} = 50\hat{i} + 100\hat{j}$ m/s.
Answer: A ($50\hat{i} + 100\hat{j}$)
Solution
The midpoint is at distance $r = 4$ cm $= 0.04$ m from each wire. Field due to one long straight wire:
$$B = \frac{\mu_0}{4\pi}\cdot\frac{2I}{r} = 10^{-7}\times\frac{2\times 30}{0.04} = 10^{-7}\times 1500 = 1.5\times10^{-4}\ \text{T}$$The currents are antiparallel, so at the midpoint both fields point the same way and add:
$$B_{\text{net}} = 2B = 3\times10^{-4}\ \text{T} = 300\ \mu\text{T}$$Answer: A ($300$)
Solution
The magnetic force $\vec{F} = q\vec{v}\times\vec{B}$ is always perpendicular to $\vec{B}$. Since $\vec{a} = \vec{F}/m$, the acceleration must also be perpendicular to $\vec{B}$:
$$\vec{a}\cdot\vec{B} = 0$$$$\left(4\hat{i} - \frac{x}{2}\hat{j}\right)\cdot(3\hat{i} + 2\hat{j}) = 0$$$$4\times 3 + \left(-\frac{x}{2}\right)\times 2 = 12 - x = 0$$Hence $x = 12$.
Answer: 12
Solution
By the work-energy theorem, the change in kinetic energy equals the total work done. The magnetic force does no work (it is always perpendicular to velocity), so only the electric force contributes:
$$\Delta KE = W_E = \int \vec{F}_E\cdot d\vec{r} = \int_{x=0}^{x=L} qE_0 e^{-\lambda x}\,dx$$Only the $x$-displacement matters since $\vec{E}$ points along $\hat{x}$. Integrating:
$$\Delta KE = qE_0\left[\frac{-e^{-\lambda x}}{\lambda}\right]_0^{L} = \frac{qE_0}{\lambda}\left[1 - e^{-\lambda L}\right]$$Answer: A ($\dfrac{q E_0}{\lambda}\left[1 - e^{-\lambda L}\right]$)
Solution
The magnetic moment is $\vec{m} = IA\,\hat{n}$ with $A = \pi r^2 = \pi(0.02)^2 = 4\pi\times10^{-4}\ \text{m}^2$.
$$\vec{m} = 100\sqrt{2}\times 4\pi\times10^{-4}\times\frac{\hat{k}+\hat{i}}{\sqrt{2}} = 100\times 4\pi\times10^{-4}(\hat{i}+\hat{k}) = 0.04\pi(\hat{i}+\hat{k})$$The torque $\vec{\tau} = \vec{m}\times\vec{B}$ with $\vec{B} = 4\times10^{-3}(3\hat{i}+2\hat{k}) = (0.012\hat{i} + 0.008\hat{k})$:
$$\vec{\tau} = 0.04\pi(\hat{i}+\hat{k})\times(0.012\hat{i}+0.008\hat{k})$$$$\hat{i}\times\hat{k} = -\hat{j},\quad \hat{k}\times\hat{i} = \hat{j}$$$$\vec{\tau} = 0.04\pi\big[0.008(\hat{i}\times\hat{k}) + 0.012(\hat{k}\times\hat{i})\big] = 0.04\pi\big[-0.008\hat{j} + 0.012\hat{j}\big]$$$$\vec{\tau} = 0.04\pi\times 0.004\,\hat{j} = 1.6\times10^{-4}\pi\,\hat{j} = 5.024\times10^{-4}\,\hat{j} = 5024\times10^{-7}\,\hat{j}$$Answer: D ($5024 \times 10^{-7}\,\hat{j}$)
Solution
The velocity component along $\hat{k}$ (parallel to $\vec{B}$) is unaffected by the field and produces the drift along the axis; the perpendicular component causes circular motion, giving helical motion.
Period of one revolution:
$$T = \frac{2\pi m}{qB} = \frac{2\pi\times 5\times10^{-6}}{5\pi\times10^{-6}\times 0.1} = \frac{2\pi\times 5\times10^{-6}}{5\pi\times10^{-7}} = 20\ \text{s}$$Pitch (axial distance per revolution) using $v_z = 2\times10^{-2}$ m/s:
$$\text{pitch} = v_z\,T = 2\times10^{-2}\times 20 = 0.4\ \text{m}$$Distance along $\hat{k}$ in 5 revolutions:
$$\alpha = 5\times 0.4 = 2\ \text{m}$$Answer: 2