$A$ metal sample carrying a current along the $x-$ axis with current density $J$ is subjected to a magnetic field $B$ (along the $z-$ axis). The electric field $E$ developed along the $y-$ axis is directly proportional to $J$ as well as $B$. The constant of proportionality has the $SI$ unit:

What is the behavior of a charged particle moving in a region where the electric field $\vec{E}$ and magnetic field $\vec{B}$ are perpendicular to each other?

The magnetic force depends on $v$,which depends on the inertial frame of reference. Does the magnetic force differ from one inertial frame to another? Is it reasonable that the net acceleration has a different value in different frames of reference?

In the product $\overrightarrow{F} = q(\vec{v} \times \overrightarrow{B})$ where $\overrightarrow{B} = B \hat{i} + B \hat{j} + B_{0} \hat{k}$. Given $q = 1$,$\vec{v} = 2 \hat{i} + 4 \hat{j} + 6 \hat{k}$,and $\overrightarrow{F} = 4 \hat{i} - 20 \hat{j} + 12 \hat{k}$,what is the complete expression for $\overrightarrow{B}$?

$A$ charged particle carrying charge $1\,\mu C$ is moving with velocity $(2 \hat{i} + 3 \hat{j} + 4 \hat{k})\, ms^{-1}$. If an external magnetic field of $(5 \hat{i} + 3 \hat{j} - 6 \hat{k}) \times 10^{-3}\, T$ exists in the region where the particle is moving,then the force on the particle is $\overrightarrow{F} \times 10^{-9}\, N$. The vector $\overrightarrow{F}$ is:

$A$ 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