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$
${15}$
${30}$
${45}$
${60}$
A collimated beam of charged and uncharged particles is directed towards a hole marked $P$ on a screen as shown below. If the electric and magnetic fields as indicated below are turned $ON$
A particle with charge $q$, moving with a momentum $p$, enters a uniform magnetic field normally. The magnetic field has magnitude $B$ and is confined to a region of width $d$, where $d < \frac{p}{{Bq}}$, The particle is deflected by an angle $\theta $ in crossing the field
A rectangular region $A B C D$ contains a uniform magnetic field $B_0$ directed perpendicular to the plane of the rectangle. A narrow stream of charged particles moving perpendicularly to the side $AB$ enters this region and is ejected through the adjacent side $B C$ suffering a deflection through $30^{\circ}$. In order to increase this deflection to $60^{\circ}$, the magnetic field has to be
When a charged particle moving with velocity $\vec v$ is subjected to a magnetic field of induction $\vec B$, the force on it is non-zero. This implies that
The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to