$A$ charged particle is moving in a uniform magnetic field in a circular path of radius $R$. When the kinetic energy of the particle is increased to three times,then the new radius will be

$(a)$ $A$ 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?

When a charged particle enters a uniform magnetic field,its kinetic energy

$A$ 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:

$A$ particle of charge $q$,mass $m$ enters a region of magnetic field $B$ with velocity $v_0 \widehat{i}$. Find the value of $d$ if the particle emerges from the region of magnetic field at an angle $30^{\circ}$ to its initial velocity:-

$A$ charged particle enters a uniform magnetic field perpendicular to its velocity. The magnetic field: