At $t$ = $0$, a positively charged particle of mass $m$ is projected from the origin with velocity $u_0$ at an angle $37^o $ from the $x-$axis as shown in the figure. A constant magnetic field ${\vec B_0} = {B_0}\hat j$ is present in space. After a time interval $t_0$ velocity of particle may be:-
${u_0}\left[ {\frac{{\sqrt {39} }}{2}\hat i + \frac{3}{5}\hat j - \frac{{\hat k}}{{\sqrt 3 }}} \right]$
${u_0}\left[ {\frac{{\hat i}}{{\sqrt 3 }} + \frac{{\hat j}}{{\sqrt 2 }} + \frac{{\hat k}}{{\sqrt 6 }}} \right]$
${u_0}\left[ {\frac{{\sqrt {39} }}{{10}}\hat i + \frac{3}{5}\hat j + \frac{1}{4}\hat k} \right]$
${u_0}\left[ {\frac{{\sqrt {39} }}{{10}}\hat i + \frac{3}{5}\hat j + \frac{1}{2}\hat k} \right]$
Assertion : A proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal.
Reason : Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field $\overrightarrow B $ in a direction perpendicular to $\overrightarrow B $, will describe circular trajectories of equal radii.
An electron enters a magnetic field whose direction is perpendicular to the velocity of the electron. Then
An $\alpha$-particle (mass $4 amu$ ) and a singly charged sulfur ion (mass $32 amu$ ) are initially at rest. They are accelerated through a potential $V$ and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_5$, respectively. The ratio $\left(r_s / r_\alpha\right)$ is. . . . .$(4)$
Derived force on moving charge in uniform magnetic field with velocity $\overrightarrow {{v_d}} $.
An electron is moving in the north direction. It experiences a force in vertically upward direction. The magnetic field at the position of the electron is in the direction of