Question

A proton of magnetic field $0.2T$ enters a magnetic field perpendicular with a velocity $=6.0\times {10}^{5}m/sec$. Calculate the acceleration of the proton and radius of the path.

Answer 1

Given,
Magnetic field $B=0.2T$
Speed of proton $v=6.0\times {10}^{5}m/s$
Charge $q=e=1.6\times {10}^{-19}C$
$\theta ={90}^o$
$\therefore$ $F=qvB\sin \theta$
$F=1.6\times {!0}^{-19}\times 6.0\times {!0}^5\times 0.2\times \sin {90}^o$
$F=1.92\times {10}^{-14}N$
$\because$ $F=ma$
$\therefore$ Acceleration $a=\frac{F}{m}=\frac{1.92\times {10}^{-14}}{1.67\times {10}^{-27}}$

$a=1.15\times {10}^{13}m/S^2$

Radius of path $r=\frac{mv}{qB}=\frac{1.67\times {10}^{-27}\times 6.0\times {10}^5}{1.6\times {10}^{-19}\times 0.2}$
$r=0.031m$