$A$ particle of mass $m$ and charge $q$ is in an electric and magnetic field given by $\vec E = 2\hat i + 3\hat j$ and $\vec B = 4\hat j + 6\hat k$. The charged particle is shifted from the origin $(0, 0, 0)$ to the point $P(1, 1, 0)$ along a straight path. The magnitude of the total work done is: (in $q$)

$A$ proton moves with a velocity of $5 \times 10^6 \hat{j} \text{ ms}^{-1}$ through a uniform electric field $\vec{E} = 4 \times 10^6 [2 \hat{i} + 0.2 \hat{j} + 0.1 \hat{k}] \text{ Vm}^{-1}$ and a uniform magnetic field $\vec{B} = 0.2 [\hat{i} + 0.2 \hat{j} + \hat{k}] \text{ T}$. The approximate net force acting on the proton is

$A$ charge '$q$' moves with velocity '$\overrightarrow{v}$' through an electric field '$\overrightarrow{E}$' as well as a magnetic field '$\overrightarrow{B}$'. Then the force acting on it is:

An electron enters an electric field with intensity $\vec{E} = 3\hat{i} + 6\hat{j} + 2\hat{k} \text{ V m}^{-1}$ and a magnetic field with induction $\vec{B} = 2\hat{i} + 3\hat{j} \text{ T}$ with a velocity $\vec{v} = 2\hat{i} + 3\hat{j} \text{ m s}^{-1}$. The magnitude of the force acting on the electron is (Given,$e = -1.6 \times 10^{-19} \text{ C}$)

$A$ charge $q$ is released in the presence of an electric field $(E)$ and a magnetic field $(B)$. After some time,its velocity is $v$. Then:

$A$ charge $q$ moves in a region where both an electric field and a magnetic field exist. The total force on it is: