Question

A charged particle having charge $10^{-6}C$ and mass of $10^{-10}kg$ is fired the middle of the plate making an angle $30^{\circ }$ with plane of the plate. Length of the plate is $0.17m$ and it is separated by $0.1m$. Electric field $E=10^{-3}N/C$ is present between the plates. Just outside the plates magnetic field is present. Find the magnitude of the magnetic field (in $mT$ ) of the charged particle perpendicular to the plane of the figure, if it has to graze the plate at $C$ and $A$ parallel to the surface of the plate (neglect gravity).

Answer 1

Answer: 3.46

To graze at $C$
Using equation of trajectory of parabola,
$y=x\tan \theta -\frac{a{x}^2}{2v^2{\cos }^2\theta }$....(i)
Hre, $a=\frac{qE}{m}=\frac{10^{-6}\times 10^{-3}}{10^{-10}}=10m/s^2$
Substituting in Eq. (i), we have
$0.05=0.17\tan 30^{\circ }-\frac{10\times (0.17{)}^2}{2v^2\times (\sqrt{3}/2{)}^2}$
Solving this equation, we have $v=2m/s$
In magnetic field, $AC=2r$
or $0.1=2r$
or $r=0.05m=\frac{mv\cos 30^{\circ }}{Bq}$
$\therefore B=\frac{mv\cos 30^{\circ }}{(0.05)q}=\frac{\left(10^{-10}\right)(2)(\sqrt{3}/2)}{(0.05)\left(10^{-6}\right)}$
$=3.46\times 10^{-3}T=3.46mT$