When a charged particle in motion enters in a uniform magnetic field perpendicularly then its

For a plane electromagnetic wave, the magnetic field at a point $\mathrm{x}$ and time $\mathrm{t}$ is :

$\overrightarrow{\mathrm{B}}\left(\mathrm{x},\mathrm{t}\right)=\left[1.2\times {10}^{-7}\sin \left(0.5\times {10}^3\mathrm{x}+1.5\times {10}^{11}\mathrm{t}\right)\hat{\mathrm{k}}\right] \mathrm{T}$.

The instantaneous electric field $\overrightarrow{\mathrm{E}}$ corresponding to $\overrightarrow{\mathrm{B}}$ is :

Two magnetic dipoles $\mathrm{X}$ and $\mathrm{Y}$ are placed at a separation $\mathrm{d}$ , with their axes perpendicular to each other. The dipole moment of $\mathrm{Y}$ is twice that of $\mathrm{X}$ . A particle of charge $\mathrm{q}$ is passing through their mid-point $\mathrm{P}$ , at angle $\mathrm{\theta }={45}^{\mathrm{o}}$ with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? ( $\mathrm{d}$ is much larger than the dimension of the dipole)

An electron enters a magnetic field of $\left(3 \hat{i}+4 \hat{j}\right) \mathrm{T}$ with a velocity of $\left(6 \hat{j}+4 \hat{k}\right) \mathrm{m} {\mathrm{s}}^{-1}.$ The acceleration produced is

($\frac{e}{m}$ of electron $=1.76\times {10}^{11} \mathrm{C} {\mathrm{kg}}^{-1}$)

Consider a negatively charged particle moving with a velocity $v$ in a magnetic field $B$ applied perpendicular to the plane of the paper (into the paper). The particle follows the path $A$ or $B$ or $C$ or $D$ (shown in the figure)