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

An electron beam passes through a magnetic field of $2 \times 10^{-3} \, Wb/m^2$ and an electric field of $1.0 \times 10^4 \, V/m$ both acting simultaneously. The path of the electron remains undeviated. What are the speed of the electron and the radius of the electron path if the electric field is removed?

An electron (mass $= 9 \times 10^{-31} \, kg$,charge $= 1.6 \times 10^{-19} \, C$) with a kinetic energy of $7.2 \times 10^{-18} \, J$ is moving in a circular orbit in a magnetic field of $9 \times 10^{-5} \, Wb/m^2$. The radius of the orbit is ..... $cm$.

$A$ 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 $^{\circ}$)

$A$ charged particle having charge $q$ is moving perpendicularly to a uniform magnetic field with linear speed $v$ on a circular path of radius $R$. The periodic time of revolution of the particle . . . . . . .

Two charged particles of mass $m$ and charge $q$ each are projected from the origin simultaneously with the same speed $V$ in a transverse magnetic field $B$. If $\vec{r}_1$ and $\vec{r}_2$ are the position vectors of the particles (with respect to the origin) at $t = \frac{\pi m}{qB}$, then the value of $\vec{r}_1 \cdot \vec{r}_2$ at that time is: