JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Magnetic Effects of Current

Between $P$ and $Q$ only the electric field does work, since the magnetic force always stays perpendicular to the velocity. That work, $qE(2a)$, equals the change in kinetic energy $\frac{1}{2}m(2v{)}^2-\frac{1}{2}m{v}^2=\frac{3}{2}m{v}^2$, giving $E=\frac{3m{v}^2}{4qa}$, so statement (a) holds. The instantaneous power delivered at $P$ is $qEv=\frac{3m{v}^3}{4a}$, confirming statement (b). At $Q$ the velocity $-2v\hat{j}$ is perpendicular to $E\hat{i}$, so the electric field does no work there and the magnetic field never does, which makes statement (c) correct. For the angular momentum, at $P$ the position $a\hat{i}$ is parallel to the velocity $v\hat{i}$ so ${\vec{L}}_P=0$, whereas at $Q$, ${\vec{L}}_Q=m(-2v\hat{j})\times (2a\hat{i})=4mva\hat{k}$; the difference in magnitudes is $4mva$, not $2mva$, so statement (d) fails. The correct statements are (a), (b) and (c).