Question 6

Given $q = 10^{-6}$ C, $m = 10^{-6}$ kg.

From $t=0$ to $0.2$ s (Electric field is ON): $a_x = \frac{qE}{m} = \frac{10^{-6} \times 1}{10^{-6}} = 1 \text{ ms}^{-2}$ $a_y = -10 \text{ ms}^{-2}$ (due to gravity)

At $t=0.2$ s: $v_x = 1 + 1(0.2) = 1.2 \text{ ms}^{-1}$ $v_y = 2 - 10(0.2) = 0 \text{ ms}^{-1}$ Vertical displacement: $y = 2(0.2) - \frac{1}{2}(10)(0.2)^2 = 0.4 - 0.2 = 0.2 \text{ m}$ ($20$ cm)

For $t > 0.2$ s (E is OFF, $B = 6 \hat{j}$ T is ON): The particle is in a uniform magnetic field with velocity perpendicular to $B$, so it moves in a circular path in the $XZ$ plane. Gravity continues to act vertically downwards ($a_y = -10 \text{ ms}^{-2}$).

Let's analyze the vertical distance at given times (where $\Delta t = t - 0.2$):

• At $t=0.3$ s ($\Delta t = 0.1$ s): $y = 0.2 + 0(0.1) - \frac{1}{2}(10)(0.1)^2 = 0.2 - 0.05 = 0.15 \text{ m} = 15 \text{ cm}$. (So (A) is correct)
• At $t=0.4$ s ($\Delta t = 0.2$ s): $y = 0.2 + 0 - \frac{1}{2}(10)(0.2)^2 = 0.2 - 0.2 = 0 \text{ cm}$. (So (B) is incorrect)

Radius of the circular path in the $XZ$ plane: $r = \frac{mv_{\perp}}{qB} = \frac{10^{-6} \times 1.2}{10^{-6} \times 6} = 0.2 \text{ m} = 20 \text{ cm}$. (So (C) is correct)

The particle hits the $XZ$ plane ($y=0$) when $t=0.4$ s, not $0.35$ s. (So (D) is incorrect)

Correct Options: (A), (C)