A charged particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by the path described by the particle is proportional to

An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5\, R)$ respectively, in a uniform magnetic field ${\rm{\vec B = }}{{\rm{B}}_0}{\rm{\hat i}}$ , each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles ?

A proton and a deutron ( $\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is:

A block of mass $m$ $\&$ charge $q$ is released on a long smooth inclined plane magnetic field $B$ is constant, uniform, horizontal and parallel to surface as shown. Find the time from start when block loses contact with the surface.

A magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{\mathrm{j}}$ exists in the region $\mathrm{a} < \mathrm{x} < 2 \mathrm{a}$ and $\vec{B}=-B_0 \hat{j}$, in the region $2 \mathrm{a} < \mathrm{x} < 3 \mathrm{a}$, where $\mathrm{B}_0$ is a positive constant. $\mathrm{A}$ positive point charge moving with a velocity $\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\dot{i}}$, where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

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