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

Derive the expression for the magnetic force acting on a moving charge $q$ with velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$.

An electron enters a region where the electrostatic field is $20\,N/C$ and the magnetic field is $5\,T$. If the electron passes undeflected through the region,then the velocity of the electron will be.....$m\,s^{-1}$.

What is the radius of the path of an electron (mass $m = 9 \times 10^{-31} \; kg$ and charge $q = 1.6 \times 10^{-19} \; C$) moving at a speed of $v = 3 \times 10^{7} \; m/s$ in a magnetic field of $B = 6 \times 10^{-4} \; T$ perpendicular to it? What is its frequency? Calculate its energy in $keV$. (Given: $1 \; eV = 1.6 \times 10^{-19} \; J$)

An electron moving with a velocity $\vec{V_1} = 2\,\hat{i}\,\text{m/s}$ at a point in a magnetic field experiences a force $\vec{F_1} = -2\hat{j}\,\text{N}$. If the electron is moving with a velocity $\vec{V_2} = 2\,\hat{j}\,\text{m/s}$ at the same point,it experiences a force $\vec{F_2} = +2\,\hat{i}\,\text{N}$. The force the electron would experience if it were moving with a velocity $\vec{V_3} = 2\hat{k}\,\text{m/s}$ at the same point is

If a proton of kinetic energy $8.35 \text{ MeV}$ enters a uniform magnetic field of $10 \text{ T}$ at right angles to the direction of the field,then the force acting on the proton is (Mass of proton $= 1.67 \times 10^{-27} \text{ kg}$ and charge of proton $= 1.6 \times 10^{-19} \text{ C}$)