Derived force on moving charge in uniform magnetic field with velocity $\overrightarrow {{v_d}} $.

A charge $q$ is released in presence of electric $(E)$ and magnetic field $(B)$ then after some time its velocity is $v$ then

A proton and an alpha particle both enter a region of uniform magnetic field $B,$ moving at right angles to the field $B.$ If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is $1\,\, MeV,$ the energy acquired by the alpha particle will be……$MeV$

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

Answer the following questions:

$(a)$ A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

$(b)$ A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

$(c)$ An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.