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 $0.17\, m$ and it is separated by $0.1\, m$. 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

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