Question

A beam of protons with velocity $4 \times 10^{5}\, m / s$ enters a uniform magnetic field of $0.3\, T$ at an angle of $60^{\circ}$ to the magnetic field. Find the radius of the helical path taken by the proton beam.

• A. 0.2 cm
• B. 1.2 cm
• C. 2.2 cm
• D. 0.122 cm

Answer 1

Answer: (B)

When the charged particle is moving at an angle to the field (other than $0^{\circ}, 90^{\circ}$ and $180^{\circ}$ ),
in this situation resolving the velocity of the particle along and perpendicular to the field, we find that the particle moves with constant velocity $v \cos \theta$ along the field and at the same time it is also moving with velocity $v \sin 6$ perpendicular to the field due to which it will describe a circle (in a plane perpendicular to the field) of radius
$r=\frac{m(v \sin \theta)}{q B}$
Here, $m=1.67 \times 10^{-27} kg$
$v=4 \times 10^{5} \,m / s$
$\theta=60^{\circ}$
$q=1.6 \times 10^{-19} \,C$
$B=0.3 \,T$
$\therefore r=\frac{1.67 \times 10^{-27} \times 4 \times 10^{5} \times(\sqrt{3} / 2)}{1.6 \times 10^{-19} \times 0.3}$
$=0.012 \,m$
$=1.2 \,cm$