$\vec E = 2\hat i + 3\hat j ;\, B = 4\hat j + 6\hat k$
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
$ = (2q\hat i + 3q\hat i) + q(\vec v \times \vec B)$
$W = {{\vec F}_{net}} \cdot \vec S$
$ = 2q + 3q$
$ = 5q$
Reason : Cyclotron frequency depends upon the velocity


