A proton accelerated by a potential difference $500\;KV$ moves though a transverse magnetic field of $0.51\;T$ as shown in figure. The angle $\theta $through which the proton deviates from the initial direction of its motion is......$^o$

An electron is moving on a circular path of radius $r$ with speed $v$ in a transverse magnetic field $B$. $e/m$ for it will be

An electron is travelling along the $x$-direction. It encounters a magnetic field in the $y$-direction. Its subsequent motion will be

An electron is moving along the positive $X$-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$-axis. This can be done by applying the magnetic field along

A charge particle moving in magnetic field $B$, has the components of velocity along $B$ as well as perpendicular to $B$. The path of the charge particle will be

A particle of charge $q$ and mass $m$ moving with a velocity $v$ along the $x$-axis enters the region $x > 0$ with uniform magnetic field $B$ along the $\hat k$ direction. The particle will penetrate in this region in the $x$-direction upto a distance $d$ equal to