Question

A beam of protons with a velocity $4.0\times {10}^{5}m{s}^{-1}$ enters a uniform magnetic field of 0.3 T at an angle of ${60}^o$ to the magnetic field. If the radius of the helical path taken by the proton beam is $1.2\times {10}^{-X}m$. Find X?

Answer 1

The radius of helical path:

$r=\frac{m{v}_{\perp }}{Bq}$
$=\frac{mvsin\theta }{Bq}=\frac{(1.67\times {10}^{-27})(4\times {10}^5)\left(sin{60}^o\right)}{\left(0.3\right)(1.6\times {10}^{-19})}$
$=1.2\times {10}^{-2}m$