Question

A proton accelerated by a potential difference $500\text{kV}$ flies through a uniform transverse magnetic field with induction $B=0.51\text{T}$. The field occupies a region of space $d=10\text{cm}$ thickness as in figure. Find the angle (in degree) through which the proton deviates from the initial direction of motion.
Take charge of proton as $1.6\times {10}^{-19}\text{C}$ and mass of the proton is $1.6\times {10}^{-27}\text{kg}$

Answer 1

If $v$ is the speed of the proton, then
$\frac{1}{2}m{v}^2=qVv=\sqrt{\frac{2qV}{m}}$
Here $q$ and $m$ are the charge and mass of the proton respectively.
The proton traverses circular path of radius $R$ in the magnetic field, where
$R=\frac{mv}{qB}$


From the figure,

$\sin \alpha =\frac{d}{R}=\frac{d}{\frac{m\left(\sqrt{\frac{2qV}{m}}\right)}{qB}}=Bd\sqrt{\frac{q}{2mV}}$
$\sin \alpha =0.51\times 10\times {10}^{-2}\sqrt{\frac{1.6\times {10}^{-19}}{2\times 1.67\times {10}^{-27}\times 500\times {10}^3}}\approx 0.5$
On substituting the values,
we get $\alpha ={30}^{\circ }$

Why This Question? Key Concepts: Magnetic force due to transverse magnetic field acts as centripetal force for the motion of particle, when $\overrightarrow{v}$ is perpendicular to $\overrightarrow{B}.$