A proton and an $\alpha -$ particle (with their masses in the ratio of $1 : 4$ and charges in the ratio of $1:2$ are accelerated from rest through a potential difference $V$. If a uniform magnetic field $(B)$ is set up perpendicular to their velocities, the ratio of the radii $r_p : r_{\alpha }$ of the circular paths described by them will be

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

If the magnetic field is parallel to the positive $y-$axis and the charged particle is moving along the positive $x-$axis (Figure), which way would the Lorentz force be for

$(a)$ an electron (negative charge),

$(b)$ a proton (positive charge).

If the direction of the initial velocity of the charged particle is neither along nor perpendicular to that of the magnetic field, then the orbit will be

A magnetic field