Question

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 $=1.67\times {10}^{-27}kg$

• A. 2.3 cm
• B. 5.35 cm
• C. 4.35 cm
• D. 6.35 cm

Answer 1

The correct option is C 4.35 cm

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
$r=\frac{m\left(v\sin \theta \right)}{qB}=\frac{(1.67\times {10}^{-27})\times (4\times {10}^5\times \sin {60}^0)}{1.6\times {10}^{-19}\times 0.3}=\frac{(1.67\times {10}^{-27})\times (4\times {10}^5\times \frac{\sqrt{3}}{2})}{1.6\times {10}^{-19}\times 0.3}=\frac{2\times {10}^{-2}}{\sqrt{3}}$
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: $P=v\cos \theta T=(4\times {10}^5)\times \cos {60}^0\times \frac{2\pi }{3}\times {10}^{-7}=\frac{4\pi }{3}\times {10}^{-2}=4.35\times {10}^{-2}m$