Question

There is a constant homogeneous electric field of $100{\text{Vm}}^{–1}$ within the region $x=0$ and $x=0.167\text{m}$ pointing in the positive $x-$direction. There is a constant homogeneous magnetic field $B$ within the region $x=0.167\text{m}$ and $x=0.334\text{m}$ pointing in the $z-$direction. A proton (at rest) at the origin $(x=0,y=0)$ is released in the positive $x-$direction. The minimum strength of the magnetic field $B$(in $mT$), so that the proton will reach $x=0,y=0.167\text{m}$ is
(mass of the proton $=1.67\times {10}^{–27}\text{kg}$)

Answer 1

The situation described in the problem is shown in figure, as electric field is along $x-$axis, so proton will be accelerated by the electric field and will enter the magnetic field at A(i.e., $x=0.167,y=0$) with velocity $v$ along x-axis such that

$\frac{1}{2}m{v}^2=W=Fd=qEd$
$v={\left[\frac{2qEd}{m}\right]}^{1/2}$
$v={\left[\frac{2\times 1.6\times {10}^{-19}\times 100\times 0.167}{1.67\times {10}^{-27}}\right]}^{1/2}$
$v=4\sqrt{2}\times {10}^4\text{m/s}$

Now, as proton is moving perpendicular to magnetic field, it will describe a circular path in the magnetic field with radius $r,$ such that
$r=\frac{mv}{qB}$

And as it comes back at $C[x=0;y=0.167\text{m}]$ its path in the magnetic field will be a semicircle such that
$y=2r=\frac{2mv}{qB}$
$B=\frac{2mv}{qy}$
$B=\frac{2\times 1.67\times {10}^{-27}\times 4\sqrt{2}\times {10}^4}{1.6\times {10}^{-19}\times 0.167}$
$B=\frac{1}{\sqrt{2}}\times {10}^{-2}$
$B=7.07\text{mT}$