Magnetic Effects of Current JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Magnetic Effects of Current, free to read — no sign-in needed. The full chapter has 47 questions; sign in to attempt the remaining 44 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   In a vacuum chamber, a particle of charge $1\ \mu\text{C}$ and mass $1$ mg is projected with a velocity $(\hat{i} + 2\hat{j})$ ms$^{-1}$ from the $XZ$ plane at time $t = 0$ in an electric field of $1\hat{i}$ Vm$^{-1}$. At $t = 0.2$ s, the electric field is switched off and a magnetic field of $6\hat{j}$ T is switched on. The acceleration due to gravity is $-10\hat{j}$ ms$^{-2}$. Correct option(s) is/are:
   1. A.The vertical distance of the particle from the $XZ$ plane at $t = 0.3$ s is $15$ cm.
   2. B.The vertical distance of the particle from the $XZ$ plane at $t = 0.4$ s is $10$ cm.
   3. C.The radius of the trajectory of the particle for $t \gt 0.2$ s is $20$ cm.
   4. D.The particle will be in the $XZ$ plane at $t = 0.35$ s.
   Show answer & solution

   Answer: A,C

   For the time interval $0 \le t \le 0.2$ s, the particle is under the influence of the electric field and gravity. The charge-to-mass ratio of the particle is $\dfrac{q}{m} = \dfrac{10^{-6}}{10^{-6}} = 1$ C kg$^{-1}$. The acceleration of the particle is: $\vec{a} = \dfrac{q\vec{E}}{m} + \vec{g} = 1(\hat{i}) - 10\hat{j} = \hat{i} - 10\hat{j}$ ms$^{-2}$ Given the initial velocity $\vec{v}_0 = \hat{i} + 2\hat{j}$ ms$^{-1}$ and initial vertical position $y_0 = 0$ (since it is projected from the $XZ$ plane), we can find the velocity and position at $t = 0.2$ s. Velocity at $t = 0.2$ s: $\vec{v}(0.2) = \vec{v}_0 + \vec{a}t = (\hat{i} + 2\hat{j}) + (\hat{i} - 10\hat{j})(0.2) = 1.2\hat{i}$ ms$^{-1}$ Vertical position at $t = 0.2$ s: $y(0.2) = y_0 + v_{0y}t + \dfrac{1}{2}a_y t^2 = 0 + 2(0.2) - \dfrac{1}{2}(10)(0.2)^2 = 0.4 - 0.2 = 0.2$ m $= 20$ cm For $t \gt 0.2$ s, the electric field is switched off and the magnetic field $\vec{B} = 6\hat{j}$ T is switched on. The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ acts only in the $XZ$ plane because $\vec{B}$ is along the $y$-axis. The vertical motion is only affected by gravity. Let $t' = t - 0.2$ be the time elapsed after the fields are switched. The initial vertical velocity for this phase is $v_y(0.2) = 0$. The vertical position as a function of $t'$ is: $y(t') = y(0.2) + v_y(0.2)t' - \dfrac{1}{2}gt'^2 = 0.2 - 5t'^2$ Evaluating the options: At $t = 0.3$ s ($t' = 0.1$ s): $y(0.1) = 0.2 - 5(0.1)^2 = 0.2 - 0.05 = 0.15$ m $= 15$ cm. (Option A is correct) At $t = 0.4$ s ($t' = 0.2$ s): $y(0.2) = 0.2 - 5(0.2)^2 = 0.2 - 0.2 = 0$ m $= 0$ cm. (Option B is incorrect) At $t = 0.35$ s ($t' = 0.15$ s): $y(0.15) = 0.2 - 5(0.15)^2 = 0.2 - 0.1125 = 0.0875$ m $\neq 0$. (Option D is incorrect) For the radius of the trajectory for $t \gt 0.2$ s, the particle moves in a helical path. The radius of the circular projection in the $XZ$ plane is determined by the velocity perpendicular to the magnetic field, $v_{\perp} = 1.2$ ms$^{-1}$. $R = \dfrac{mv_{\perp}}{qB} = \dfrac{10^{-6} \times 1.2}{10^{-6} \times 6} = 0.2$ m $= 20$ cm. (Option C is correct) Answer: The vertical distance of the particle from the $XZ$ plane at $t = 0.3$ s is $15$ cm.; The radius of the trajectory of the particle for $t \gt 0.2$ s is $20$ cm.
2. Q2JEE Advanced Adv 2021 (Paper 1)
   An $\alpha$-particle (mass $4$ amu) and a singly charged Sulphur ion (mass $32$ amu) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulphur ion move in circular orbits of radii ${R}_{\alpha }$ and ${R}_{S}$, respectively. The ratio $\left(\dfrac{{R}_{S}}{{R}_{\alpha }}\right)$ is __________ .
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   Answer: 4

   When a particle moves on a circular path in a magnetic field then, $∵R=\dfrac{mv}{qB}=\dfrac{\sqrt{2mE}}{qB}=\dfrac{\sqrt{2mqV}}{qB}=\dfrac{\sqrt{2mV}}{\sqrt{q}B}$ According to given data, $∵V=\text{constant} \text{\&}B=\text{constant}$ $∴R\propto \sqrt{\dfrac{m}{q}}$ $\dfrac{{R}_{\alpha }}{{R}_{S}}=\sqrt{\dfrac{{m}_{\alpha }}{{q}_{\alpha }}\times \dfrac{{q}_{S}}{{m}_{S}}}=\sqrt{\dfrac{4}{2}\times \dfrac{1}{32}}$ $\Rightarrow \dfrac{{R}_{\alpha }}{{R}_{S}}=\sqrt{\dfrac{1}{16}}=\dfrac{1}{4}$ Then, $\dfrac{{R}_{S}}{{R}_{\alpha }}=4$
3. Q3JEE Advanced Adv 2008 (Paper 2)
   Statement 1 The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. and Statement 2 Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized.
   1. A.Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
   2. B.Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
   3. C.Statement 1 is true, Statement 2 is false.
   4. D.Statement 1 is false, Statement 2 is true
   Show answer & solution

   Answer: (C)

   $c \phi=B I N A$ $$\therefore \phi=\left(\frac{B N A}{c}\right) I$$ Using iron core, value of magnetic field increases. So, deflection increases for same current. Hence, sensitivity increases. Soft iron can be easily magnetized or demagnetized. $\therefore$ correct option is (c).