Question

A beam of protons with a velocity $4\times 10^{5}m{s}^{-1}$ 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 = $1.67\times 10^{-27}kg$

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

Answer 1

Answer: (B)

$\frac{m(vsin\theta {)}^2}{r}=q(vsin\theta )$ B
$r=\frac{mvsin\theta }{qB}=\frac{1.67\times 10^{-27}\times 4\times 10^5\times sin60^{\circ }}{1.6\times 10^{-19}\times 0.3}$ = 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 = $(vcos\theta )$ T = $4\times 10^5\times cos60^{\circ }\times 2.19\times 10^{-7}=4.38\times 10^{-2}m$ = 4.38 cm