An electron and a proton have equal kinetic energies. They enter a magnetic field perpendicularly. Then:

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path.
Reason $(R) :$ The magnetic field in that region is along the direction of velocity of the electron.
In the light of the above statements,choose the correct answer from the options given below.

An electron $(e^-)$ is moving in the North direction and the magnetic field at this place is in the upward direction. Then,the electron $(e^-)$ will be deviated towards:

$(a)$ $A$ 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?

$A$ charged particle, when entering a uniform magnetic field, moves in a helical path. If its angular velocity is $4 \pi \times 10^6 \text{ rad s}^{-1}$ and its velocity in the direction of the magnetic field is $3 \times 10^5 \text{ m s}^{-1}$, then the pitch of the helix is: (in $\text{ cm}$)

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The angle of incidence is $45^{\circ}$. The radius of the circular portion of the path is: