JEE Advanced 2026 — Paper 2 — PHY-II solution

JEE Advanced2026Paper 2PHY-II

In a vacuum chamber, a particle of charge $1 μ$ C and mass $1$ mg is projected with a velocity $(\hat{i} + 2 \hat{j})$ ms $^{- 1}$ from the $X Z$ 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:

A
The vertical distance of the particle from the $X Z$ plane at $t = 0.3$ s is $15$ cm.
B
The vertical distance of the particle from the $X Z$ plane at $t = 0.4$ s is $10$ cm.
C
The radius of the trajectory of the particle for $t > 0.2$ s is $20$ cm.
D
The particle will be in the $X Z$ plane at $t = 0.35$ s.

Solution

Phase 1: $0 \leq t \leq 0.2$ s. With $q = 1 0^{- 6}$ C and $m = 1 0^{- 6}$ kg, $q E / m = \hat{i}$ m/s $^{2}$ and gravity contributes $- 10 \hat{j}$ m/s $^{2}$ . So $a_{1} = \hat{i} - 10 \hat{j}$ m/s $^{2}$ .

At $t = 0.2$ s, $v = (1 + 0.2) \hat{i} + (2 - 2) \hat{j} = 1.2 \hat{i}$ m/s, and $y = 2 (0.2) - \frac{1}{2} (10) (0.2)^{2} = 0.2$ m.

Phase 2: $t > 0.2$ s. With $B = 6 \hat{j}$ T, the magnetic force on $v_{⊥} = 1.2 \hat{i}$ has no $y$ component, so the $y$ -motion is pure free fall thereafter: $y (t) = 0.2 - 5 (t - 0.2)^{2}$ .

Option (A). At $t = 0.3$ s, $y = 0.2 - 5 (0.1)^{2} = 0.15$ m $= 15$ cm. Correct.

Option (B). At $t = 0.4$ s, $y = 0.2 - 5 (0.2)^{2} = 0$ . Distance is $0$ cm, not $10$ cm. Incorrect.

Option (C). The in-plane circular motion in the $X Z$ plane has radius $R = \frac{m v _{⊥}}{q B} = \frac{( 1 0 ^{- 6} ) ( 1.2 )}{( 1 0 ^{- 6} ) ( 6 )} = 0.2$ m $= 20$ cm. Correct (interpreting trajectory radius as the helix radius).

Option (D). $y = 0$ when $5 (t - 0.2)^{2} = 0.2 \Rightarrow t - 0.2 = 0.2 \Rightarrow t = 0.4$ s, not $0.35$ s. Incorrect.

$∴$ Correct options are (A) and (C).

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