Question

A beam of protons with a velocity of $4\times {10}^5\text{m/s}$ enters a uniform magnetic field of $B=0.3\text{T}.$The velocity makes an angle of ${60}^{\circ }$ with the magnetic field. Find the pitch of the helix$\text{in cm}$.
(mass of proton $m_p=1.67\times {10}^{-27}\text{kg}$)

Answer 1

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 $v\sin \theta$, where $v$ is the projected velocity.
So, the particle will move in a circle of radius, $r=\frac{m\left(v\sin \theta \right)}{qB}$
$\Rightarrow r=\frac{(1.67\times {10}^{-27})\times (4\times {10}^5)\times \sin {60}^{\circ }}{1.6\times {10}^{-19}\times 0.3}=\frac{2\times {10}^{-2}}{\sqrt{3}}$
Now, we have time period, $T=\frac{2\pi r}{v\sin \theta }$
$\Rightarrow T=\frac{2\pi \times \frac{2\times {10}^{-2}}{\sqrt{3}}}{4\times {10}^5\sin {60}^{\circ }}=\frac{2\pi }{3}\times {10}^{-7}$
So, the pitch of the helix is
$P=v\cos \theta T=4\times {10}^5\times \cos {60}^{\circ }\times \frac{2\pi }{3}\times {10}^{-7}$
$\Rightarrow P=4.35\times {10}^{-2}\text{m}=4.35\text{cm}$