Question

There is a constant homogeneous electric field of $100V{m}^{-1}$ within the region $x=0$ and $x=0.167$ m pointing in x-direction. There is a constant homogeneous magnetic field B within the region $X=0.167m$ and $x=0.334m$ pointing in the z-direction. A proton at rest at the origin is released in positive x-direction.If the minimum strength of the magnetic field B, so that the proton is detected back at $x=0,y=0.167m$ is equal to $7.07\times {10}^{-x}T$. (mass of proton $=1.67\times {10}^{-27}kg)$. Find x.

Answer 1

First of all the proton is accelerated in the electric field. Then it enters in magnetic field and describes a circular path. After that it leaves the magnetic field in negative direction. Its motion is retarded in electric field. Finally, it strikes y-axis at the same distance 0.167 m.
Let us first calculate the velocity of the proton when it enters in magnetic field after traversing in electric field.
Force acting on proton in electric field $=eE$
$\therefore$ Acceleration of proton $a=\left(\frac{eE}{m}\right)$
Using the formula $v^2+u^2=2$ as, we have
$v^2=2\left(\frac{eE}{m}\right)s=2\left(\frac{eE}{m}\right)\left(0.167\right)$ (i)
Now consider the motion in magnetic field. The proton describes a circular path of radius (AC/2).
Hence, $\frac{m{v}^2}{r}=evB$ or $B=\frac{mv}{er}$ (ii)
$\therefore B=\frac{m}{e\left(0.167/2\right)}\times \sqrt{\left[\frac{2eE}{m}\right(0.167\left)\right]}$
$=2\sqrt{(\frac{2m}{e}\times \frac{E}{0.167})}=2\sqrt{\left(\frac{2(1.67\times {10}^{-27})\left(100\right)}{(1.6\times {10}^{-19})\left(0.167\right)}\right]}$
$=\frac{{10}^{-2}}{\sqrt{2}}=7.07\times {10}^{-3}T$