$A$ positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. The particle will

$A$ particle of mass $m$,charge $Q$,and kinetic energy $K$ enters a transverse uniform magnetic field of induction $B$. After $3$ $seconds$,the kinetic energy of the particle will be .......$K$.

The magnetic force acting on a charged particle of charge $2\,\mu C$ in a magnetic field of $2\, T$ acting in the $y-$ direction,when the particle velocity is $(2\hat{i} + 3\hat{j}) \times 10^6\, m/s$ is:

$A$ charged particle of specific charge $\alpha$ is released from the origin at time $t = 0$ with velocity $\vec{V} = V_o \hat{i} + V_o \hat{j}$ in a magnetic field $\vec{B} = B_o \hat{i}$. The coordinates of the particle at time $t = \frac{\pi}{B_o \alpha}$ are (specific charge $\alpha = q/m$):

$A$ proton accelerated by a potential difference of $500 \ kV$ moves through a transverse magnetic field of $0.5 \ T$ as shown in the figure. The angle $\theta$,through which the proton deviates from the initial direction of its motion,is (mass of proton $= 1.6 \times 10^{-27} \ kg$):- (in $^{\circ}$)

The motion of a moving electron is not affected by: