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

An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2} \, T$. If the value of $e/m$ is $1.76 \times 10^{11} \, C/kg$,the frequency of revolution of the electron is

$(a)$ $A$ monoenergetic electron beam with electron speed of $5.20 \times 10^{6} \;m s^{-1}$ is subject to a magnetic field of $1.30 \times 10^{-4} \;T$ normal to the beam velocity. What is the radius of the circle traced by the beam,given $e/m$ for electron equals $1.76 \times 10^{11} \;C \;kg^{-1}$?
$(b)$ Is the formula you employ in $(a)$ valid for calculating the radius of the path of a $20 \;MeV$ electron beam? If not,in what way is it modified?

If a particle of charge $10^{-12} \, C$ moving along the $\hat{x}$ direction with a velocity $10^5 \, m/s$ experiences a force of $10^{-10} \, N$ in the $\hat{y}$ direction due to a magnetic field,then the minimum magnetic field is:

$A$ proton (mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$) enters perpendicular to a magnetic field of intensity $B = 2 \, Wb/m^2$ with a velocity $v = 3.4 \times 10^7 \, m/s$. The acceleration of the proton is:

An electron is travelling along the $x$-direction. It encounters a magnetic field in the $y$-direction. Its subsequent motion will be