Question

A beam of protons with a velocity $4\times10^5\,ms^{-1}$ enters the uniform magnetic field of 0.3T at an angle of $60^0$ to the magnetic field. Find the radius of the helical path taken by the proton beam. Also, find the pitch of the helix. Given mass of proton = $1.67\times10^{-27}\,kg$

• A. 6 cm
• B. 4.38 cm
• C. 6.34 cm
• D. 3.46 cm

Answer 1

Answer: (B)

$\frac{m(v \, sin \, \theta)^2}{r} = q(v \, sin \, \theta)$ B
$r = \frac{m\, v \,sin \, \theta}{qB} = \frac{1.67 \times 10^{-27} \times 4 \times 10^5 \times sin \, 60^\circ}{1.6 \times 10^{-19} \times 0.3} $ = 1.205 m.
T = $\frac{2\pi m}{qB} = \frac{2 \times 3.142 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.3} = 2.19 \times 10^{-7}s $
$\therefore $ Pitch = $(v \, cos \theta)$ T = $4 \times 10^5 \times cos \, 60^\circ \times 2.19 \times 10^{-7} = 4.38 \times 10^{-2}m$ = 4.38 cm