An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: mass of electron $m_e = 9 \times 10^{-31} \ kg$)

An electron falling freely under the influence of gravity enters a uniform magnetic field directed towards the south. The electron is initially deflected towards:

$A$ charged particle is moving in a magnetic field of strength $B$ perpendicular to the direction of the field. If $q$ and $m$ denote the charge and mass of the particle respectively,then the frequency of rotation of the particle 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 $m$)

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $B = B_0 \hat{k}$.

$A$ charged particle is moving with velocity $v$ in a magnetic field of induction $B$. The force on the particle will be maximum when