A beam of protons with a velocity $4\times 10^{5}m{s}^{-1}$ enters a uniform magnetic field of 0.3 T at an angle of $60^o$ to the magnetic field. Find the pitch of the helix (which is the distance travelled by a proton in the beam parallel to the magnetic field during one period of the rotation). Mass of the proton $=1.67\times 10^{-27}kg$

When a charged particle is projected at an angle $\theta$ to a magnetic field, the component of velocity parallel to the field is $v\cos \theta$ while perpendicular to the field is $v\sin \theta$, so the particle will move in a circle of radius
$r=\frac{m(v\sin \theta )}{qB}=\frac{(1.67\times 10^{-27})\times \left(4\times 10^5\times \sin 60^0\right)}{1.6\times 10^{-19}\times 0.3}=\frac{\left(1.67\times 10^{-27}\right)\times \left(4\times 10^5\times \frac{\sqrt{3}}{2}\right)}{1.6\times 10^{-19}\times 0.3}=\frac{2\times 10^{-2}}{\sqrt{3}}$
Time period: $T=\frac{2\pi r}{vsin\theta }=\frac{2\pi \times \frac{2\times 10^{-2}}{\sqrt{3}}}{4\times 10^5\times \sin 60^0}=\frac{2\pi \times \frac{2\times 10^{-2}}{\sqrt{3}}}{4\times 10^5\times \frac{\sqrt{3}}{2}}=\frac{2\pi }{3}\times 10^{-7}$
Pitch: $P=v\cos \theta T=(4\times 10^5)\times \cos 60^0\times \frac{2\pi }{3}\times 10^{-7}=\frac{4\pi }{3}\times 10^{-2}=4.35\times 10^{-2}m$