Q 12-04-219JEE MainJEE Main 2019 (11 Jan, Shift 2)Easy
A particle of mass $m$ and charge $q$ is in an electric and magnetic field given by $$\vec{E} = 2\hat{i} + 3\hat{j};\quad \vec{B} = 4\hat{j} + 6\hat{k}$$ The charged particle is shifted from the origin to the point $P(x = 1;\ y = 1)$ along a straight path. The magnitude of the total work done is:
Answer: (B) $5q$
The magnetic force is always perpendicular to the velocity and does no work. The electric force does
$$W = q\vec{E}\cdot\Delta\vec{r} = q(2\hat{i} + 3\hat{j})\cdot(\hat{i} + \hat{j}) = 5q$$
Solution by Sreeraj P, M.Sc Physics