If an electron is moving in the direction of the magnetic field $\overrightarrow{B}$ with a velocity $\overrightarrow{v}$,then the force on the electron is:

$A$ proton is moving with a uniform velocity of $10^{6} \ m/s$ along the $Y$-axis,under the joint action of a magnetic field along the $Z$-axis and an electric field of magnitude $2 \times 10^{4} \ V/m$ along the negative $X$-axis. If the electric field is switched off,the proton starts moving in a circle. The radius of the circle is nearly: (Given: $\frac{e}{m}$ ratio for proton $= 10^{8} \ C/kg$) (in $m$)

The distance between two plates is $1 \ cm$ and the potential difference is $1000 \ V$. $A$ magnetic field $B = 1 \ T$ is applied. If an electron passes through without any deflection,what will be its velocity?

$A$ proton with kinetic energy of $1 \; MeV$ moves from south to north. It experiences an acceleration of $10^{12} \; m/s^2$ due to an applied magnetic field (directed from west to east). The value of the magnetic field is: ....... $mT$ (Rest mass of proton is $1.6 \times 10^{-27} \; kg$)

An electron emitted by a heated cathode and accelerated through a potential difference of $2.0 \; kV$,enters a region with a uniform magnetic field of $0.15 \; T$. Determine the trajectory of the electron if the field:
$(a)$ is transverse to its initial velocity,
$(b)$ makes an angle of $30^{\circ}$ with the initial velocity.

An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively,in a uniform magnetic field $\vec{B} = B_0 \hat{i}$,each with an equal momentum of magnitude $P = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?