Two charged particle $A$ and $B$ each of charge $+e$ and masses $12$ $amu$ and $13$ $amu$ respectively follow a circular trajectory in chamber $X$ after the velocity selector as shown in the figure. Both particles enter the velocity selector with speed $1.5 \times 10^6 \,ms^{-1}.$ A uniform magnetic field of strength $1.0$ $T$ is maintained within the chamber $X$ and in the velocity selector.
Electric field across the conducting plate of the velocity selector is $- 10^6\, NC^{-1} \hat i$ .
Electric field across the conducting plate of the velocity selector is $10^6 \,NC^{-1} \hat i$ .
The ratio $r_A$/ $r_B$ of the radii of the circular paths for the two particles is $12/13.$
The ratio $r_A$ $/r_B$ of the radii of the circular paths for the two particles is $13/12$ .
A proton is projected with velocity $\overrightarrow{ V }=2 \hat{ i }$ in a region where magnetic field $\overrightarrow{ B }=(\hat{i}+3 \hat{j}+4 \hat{k})\; \mu T$ and electric field $\overrightarrow{ E }=10 \hat{ i } \;\mu V / m .$ Then find out the net acceleration of proton (in $m / s ^{2}$)
charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then
Two particles $x$ and $y$ have equal charges and possessing equal kinetic energy enter in a uniform magnetic field and describe circular path of radius of curvature $r_1$ and $r_2$ respectively. The ratio of their masses is
A particle with charge $-Q$ and mass m enters a magnetic field of magnitude $B,$ existing only to the right of the boundary $YZ$. The direction of the motion of the $m$ particle is perpendicular to the direction of $B.$ Let $T = 2\pi\frac{m}{{QB}}$ . The time spent by the particle in the field will be
$1$ $\mathrm{T}$ $=$ ...... Guass.