Magnetism

Concept:The particle experiences constant electric and gravitational fields initially, then only gravity and a perpendicular magnetic field, leading to helical motion.Explanation:From $t=0$ to $0.2$ s, the forces are electric ($1\hat{i}$ V/m) and gravitational ($-10\hat{j}$ m/s²).Charge $q=10^{-6}$ C, mass $m=10^{-6}$ kg, so $\frac{q}{m}=1$.Acceleration $\vec{a} = \frac{q\vec{E}}{m} + \vec{g} = 1\hat{i} -10\hat{j}$ m/s².Initial velocity $\vec{u} = \hat{i} + 2\hat{j}$ m/s.At $t=0.2$ s:$$v_x = 1 + 1(0.2) = 1.2\ \text{m/s}$$$$v_y = 2 + (-10)(0.2) = 0\ \text{m/s}$$$$y = 0 + 2(0.2) + \frac12(-10)(0.2)^2 = 0.4 - 0.2 = 0.2\ \text{m} = 20\ \text{cm}$$Velocity is $1.2\hat{i}$ m/s, position $(x_0, y=0.2$, z=?).From $t=0.2$ s onward, magnetic field $\vec{B}=6\hat{j}$ T acts.Magnetic acceleration: $\frac{q}{m}(\vec{v}\times\vec{B}) = 1 (1.2\hat{i}\times 6\hat{j}) = 7.2\hat{k}$ m/s².Gravity continues: $-10\hat{j}$ m/s².Motion is circular in the $xz$-plane with constant speed $v=1.2$ m/s, plus constant acceleration $-10\hat{j}$ in $y$.Radius: $a_c = \frac{v^2}{r} = 7.2$ ⇒ $r = \frac{1.2^2}{7.2} = 0.2$ m = 20 cm. Option C is correct.For $t>0.2$ s, $y(t) = y_0 + \frac12 a_y (t-0.2)^2 = 0.2 -5(t-0.2)^2$.At $t=0.3$ s: $y = 0.2 -5(0.1)^2 = 0.2 -0.05 = 0.15$ m = 15 cm. Option A is correct.At $t=0.4$ s: $y = 0.2 -5(0.2)^2 = 0.2 -0.2 = 0$ m. Option B (10 cm) is incorrect.At $t=0.35$ s: $y = 0.2 -5(0.15)^2 = 0.2 -0.1125 = 0.0875$ m ≠ 0. Option D is incorrect.Answer:A and C