$A$ proton is projected with velocity $\overrightarrow{V} = 2 \hat{i} \; m/s$ in a region where magnetic field $\overrightarrow{B} = (\hat{i} + 3 \hat{j} + 4 \hat{k}) \; \mu T$ and electric field $\overrightarrow{E} = 10 \hat{i} \; \mu V/m$. Find the net acceleration of the proton (in $m/s^2$).

$A$ 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

$A$ particle of mass $1 \times 10^{-26} \,kg$ and charge $1.6 \times 10^{-19} \,C$ travelling with a velocity $1.28 \times 10^6 \,ms^{-1}$ along the positive $X$-axis enters a region in which a uniform electric field $E$ and a uniform magnetic field of induction $B$ are present. If $E = -102.4 \times 10^3 \hat{k} \,NC^{-1}$ and $B = 8 \times 10^{-2} \hat{j} \,Wbm^{-2}$, the direction of motion of the particle is:

The dimension of magnetic field in $M, L, T$ and $C$ (coulomb) is given as:

$A$ particle having charge $10^{-9} \text{ C}$ moving in the $x-y$ plane in fields of $0.4 \hat{i} \text{ N/C}$ and $4 \times 10^{-3} \hat{k} \text{ T}$ experiences a force of $(4 \hat{i} + 2 \hat{j}) \times 10^{-10} \text{ N}$. The velocity of the particle at that instant is . . . . . . $\text{m/s}$.