A beam of protons with a velocity of $4\times {10}^5{\mathrm{ms}}^{-1}$  enters a uniform magnetic field of $0.3\mathrm{T}.$  The velocity makes an angle of  60° with the magnetic field. Find the radius of the helical path taken by the proton beam

Solution:

The components of the proton’s velocity along and perpendicular to the magnetic field are ${\mathrm{v}}_{\parallel }=\left(4\times {10}^5{\mathrm{ms}}^{-1}\right)\cos {60}^0=2\times {10}^5{\mathrm{ms}}^{-1}.$
 

And   ${\mathrm{v}}_{\perp }=\left(4\times {10}^5{\mathrm{ms}}^{-1}\right)\sin {60}^{\circ }=2\sqrt{3}\times {10}^5{\mathrm{ms}}^{-1}$
As the force $\mathrm{q}\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}}$ is perpendicular to the magnetic field, the component ${\mathrm{v}}_{\parallel }$ will remain constant. In the plane perpendicular to the field, the proton will describe a circle whose radius is obtained from the equation ${\mathrm{qv}}_{\perp }\mathrm{B}=\frac{{\mathrm{mv}}_{\perp }^2}{\mathrm{r}}$
Or   $\mathrm{r}=\frac{{\mathrm{mv}}_{\perp }}{\mathrm{qB}}=\frac{\left(1.67\times {10}^{-27}\mathrm{kg}\right)\times \left(2\sqrt{3}\times {10}^5{\mathrm{ms}}^{-1}\right)}{\left(1.6\times {10}^{-19}\mathrm{C}\right)\times (0.3\mathrm{T})}=0.012\mathrm{m}=1.2\mathrm{cm}$