If an electron enters a magnetic field with its velocity pointing in the same direction as the magnetic field,then

$A$ particle of mass $m$ and charge $q$ is thrown from the origin at $t = 0$ with velocity $\vec{v} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ units in a region with a uniform magnetic field $\vec{B} = 2\hat{i}$ units. After time $t = \frac{\pi m}{qB}$,an electric field $\vec{E}$ is switched on such that the particle moves in a straight line with constant speed. $\vec{E}$ may be:

$A$ particle of charge $q$ and mass $m$ is moving with a velocity $-v \hat{i} (v \neq 0)$ towards a large screen placed in the $Y-Z$ plane at a distance $d$. If there is a magnetic field $\vec{B} = B_{0} \hat{k}$,the minimum value of $v$ for which the particle will not hit the screen is

$A$ beam of ions with velocity $2 \times 10^5 \ m/s$ enters normally into a uniform magnetic field of $4 \times 10^{-2} \ T$. If the specific charge of the ion is $5 \times 10^7 \ C/kg$,then the radius of the circular path described will be $....... \ m$.

If a proton, deuteron, and $\alpha$-particle are accelerated by the same potential difference and enter perpendicular to a magnetic field, then the ratio of their kinetic energies is:

$A$ proton of velocity $v = (3 \hat{i} + 2 \hat{j}) \ m/s$ enters a magnetic field of induction $B = (2 \hat{j} + 3 \hat{k}) \ T$. The acceleration produced in the proton in $m/s^2$ is (Specific charge of proton $= 0.96 \times 10^8 \ C/kg$)