The net charge in a current carrying wire is zero still magnetic field exerts a force on it, because a magnetic field exerts force on

A particle of charge $q$ and mass $m$ is moving along the $x$ -axis with a velocity $v$ and enters a region of electric field $E$ and magnetic field $B$ as shown in figure below for which figure the net force on the charge may be zero

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

A particle of charge $ - 16 \times {10^{ - 18}}$ $coulomb$ moving with velocity $10\,\,m{s^{ - 1}}$ along the $x$-axis enters a region where a magnetic field of induction $B$ is along the $y$-axis, and an electric field of magnitude ${10^4}\,\,V/m$ is along the negative $z$-axis. If the charged particle continues moving along the $x$-axis, the magnitude of $B$ is

A particle of charge $q$ and mass $m$ starts moving from the origin under the action of an electric field $\vec E = {E_0}\hat i$ and $\vec B = {B_0}\hat i$ with velocity ${\rm{\vec v}} = {{\rm{v}}_0}\hat j$. The speed of the particle will become $2v_0$ after a time

A particle is moving in a uniform magnetic field, then