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PhyMedAdv 2026 • Paper 2

Motion of Charged Particle in Electric and Magnetic Fields

In a vacuum chamber, a particle of charge 1 μC1\ \mu\text{C} and mass 1 mg1\text{ mg} is projected with a velocity (i^+2j^) ms−1(\hat{i} + 2\hat{j})\text{ ms}^{-1} from the XZXZ plane at time t=0t = 0 in an electric field of 1i^ Vm−11\hat{i}\text{ Vm}^{-1}. At t=0.2 st = 0.2\text{ s}, the electric field is switched off and a magnetic field of 6j^ T6\hat{j}\text{ T} is switched on. The acceleration due to gravity is −10j^ ms−2-10\hat{j}\text{ ms}^{-2}. Correct option(s) is/are:

Options

A

The vertical distance of the particle from the XZXZ plane at t=0.3 st = 0.3\text{ s} is 15 cm15\text{ cm}.

Correct
B

The vertical distance of the particle from the XZXZ plane at t=0.4 st = 0.4\text{ s} is 10 cm10\text{ cm}.

C

The radius of the trajectory of the particle for t>0.2 st > 0.2\text{ s} is 20 cm20\text{ cm}.

Correct
D

The particle will be in the XZXZ plane at t=0.35 st = 0.35\text{ s}.

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric FieldKinematics

Step-by-Step Solution

Given Parameters:

  • Charge of the particle: q=1 μC=10−6 Cq = 1\ \mu\text{C} = 10^{-6}\text{ C}
  • Mass of the particle: m=1 mg=10−6 kgm = 1\text{ mg} = 10^{-6}\text{ kg}
  • Specific charge: qm=10−610−6=1 C kg−1\frac{q}{m} = \frac{10^{-6}}{10^{-6}} = 1\text{ C kg}^{-1}
  • Initial velocity at t=0t = 0: v⃗0=1i^+2j^ ms−1\vec{v}_0 = 1\hat{i} + 2\hat{j}\text{ ms}^{-1}
  • Initial position: y(0)=0y(0) = 0 (starts from the XZXZ plane)
  • Acceleration due to gravity: g⃗=−10j^ ms−2\vec{g} = -10\hat{j}\text{ ms}^{-2}

Step 1: Motion during time interval 0≤t≤0.2 s0 \le t \le 0.2\text{ s}

During this interval, an electric field E⃗=1i^ Vm−1\vec{E} = 1\hat{i}\text{ Vm}^{-1} is present along with gravity.

The electric force acting on the particle is: F⃗E=qE⃗=10−6(1i^) N\vec{F}_E = q\vec{E} = 10^{-6}(1\hat{i})\text{ N}

The net acceleration during this phase is: a⃗=qE⃗m+g⃗=(10−6×110−6)i^−10j^=(1i^−10j^) ms−2\vec{a} = \frac{q\vec{E}}{m} + \vec{g} = \left(\frac{10^{-6} \times 1}{10^{-6}}\right)\hat{i} - 10\hat{j} = (1\hat{i} - 10\hat{j})\text{ ms}^{-2}

Using kinematic equations, the velocity components at t=0.2 st = 0.2\text{ s} are: vx(0.2)=v0x+axt=1+(1)(0.2)=1.2 ms−1v_x(0.2) = v_{0x} + a_x t = 1 + (1)(0.2) = 1.2\text{ ms}^{-1} vy(0.2)=v0y+ayt=2+(−10)(0.2)=0 ms−1v_y(0.2) = v_{0y} + a_y t = 2 + (-10)(0.2) = 0\text{ ms}^{-1} vz(0.2)=0 ms−1v_z(0.2) = 0\text{ ms}^{-1}

Thus, at t=0.2 st = 0.2\text{ s}, the velocity vector is: v⃗(0.2)=1.2i^ ms−1\vec{v}(0.2) = 1.2\hat{i}\text{ ms}^{-1}

The vertical position (yy-coordinate) at t=0.2 st = 0.2\text{ s} is: y(0.2)=v0yt+12ayt2=2(0.2)+12(−10)(0.2)2=0.4−0.2=0.2 m=20 cmy(0.2) = v_{0y}t + \frac{1}{2}a_y t^2 = 2(0.2) + \frac{1}{2}(-10)(0.2)^2 = 0.4 - 0.2 = 0.2\text{ m} = 20\text{ cm}

Step 2: Motion for t>0.2 st > 0.2\text{ s}

At t=0.2 st = 0.2\text{ s}, the electric field is turned off, and a uniform magnetic field B⃗=6j^ T\vec{B} = 6\hat{j}\text{ T} is switched on.

The magnetic force acting on the particle is: F⃗B=q(v⃗×B⃗)=q[(vxi^+vyj^+vzk^)×6j^]=6q(vxk^−vzi^)\vec{F}_B = q(\vec{v} \times \vec{B}) = q\left[(v_x\hat{i} + v_y\hat{j} + v_z\hat{k}) \times 6\hat{j}\right] = 6q(v_x\hat{k} - v_z\hat{i})

Notice that F⃗B\vec{F}_B has no component along the yy-axis. Therefore, the vertical motion (yy-axis) is completely decoupled from the magnetic force and is governed solely by gravity.

Vertical Motion (yy-axis):

For t≥0.2 st \ge 0.2\text{ s}, let t′=t−0.2 st' = t - 0.2\text{ s}. The yy-acceleration is ay=−10 ms−2a_y = -10\text{ ms}^{-2}, with initial conditions y(0.2)=0.2 my(0.2) = 0.2\text{ m} and vy(0.2)=0 ms−1v_y(0.2) = 0\text{ ms}^{-1}: y(t)=y(0.2)+vy(0.2)(t−0.2)+12ay(t−0.2)2y(t) = y(0.2) + v_y(0.2)(t - 0.2) + \frac{1}{2}a_y(t - 0.2)^2 y(t)=0.2−5(t−0.2)2 my(t) = 0.2 - 5(t - 0.2)^2\text{ m}

Step 3: Verification of Options

  1. Checking Option A: At t=0.3 st = 0.3\text{ s}: y(0.3)=0.2−5(0.3−0.2)2=0.2−5(0.01)=0.2−0.05=0.15 m=15 cmy(0.3) = 0.2 - 5(0.3 - 0.2)^2 = 0.2 - 5(0.01) = 0.2 - 0.05 = 0.15\text{ m} = 15\text{ cm} Therefore, Option A is correct.

  2. Checking Option B: At t=0.4 st = 0.4\text{ s}: y(0.4)=0.2−5(0.4−0.2)2=0.2−5(0.04)=0.2−0.2=0 my(0.4) = 0.2 - 5(0.4 - 0.2)^2 = 0.2 - 5(0.04) = 0.2 - 0.2 = 0\text{ m} The vertical distance is 0 cm0\text{ cm}, not 10 cm10\text{ cm}. Therefore, Option B is incorrect.

  3. Checking Option C: For t>0.2 st > 0.2\text{ s}, the velocity component perpendicular to the magnetic field B⃗=6j^ T\vec{B} = 6\hat{j}\text{ T} is: v⊥=vx2+vz2=(1.2)2+02=1.2 ms−1v_\perp = \sqrt{v_x^2 + v_z^2} = \sqrt{(1.2)^2 + 0^2} = 1.2\text{ ms}^{-1} The radius of the circular path projected in the XZXZ plane is: R=mv⊥qB=10−6×1.210−6×6=0.2 m=20 cmR = \frac{m v_\perp}{q B} = \frac{10^{-6} \times 1.2}{10^{-6} \times 6} = 0.2\text{ m} = 20\text{ cm} Therefore, Option C is correct.

  4. Checking Option D: The particle lies in the XZXZ plane when y(t)=0y(t) = 0: 0.2−5(t−0.2)2=0  ⟹  (t−0.2)2=0.04  ⟹  t−0.2=0.2  ⟹  t=0.4 s0.2 - 5(t - 0.2)^2 = 0 \implies (t - 0.2)^2 = 0.04 \implies t - 0.2 = 0.2 \implies t = 0.4\text{ s} Thus, the particle will be in the XZXZ plane at t=0.4 st = 0.4\text{ s}, not t=0.35 st = 0.35\text{ s}. Therefore, Option D is incorrect.

Final Conclusion:

The correct statements are A and C.

Motion of Charged Particle in Electric and Magnetic Fields | Physics PYQ Solution - JEE Challenger