Question

A beam of protons with velocity $4\times {10}^{5}m/s$ enters a uniform magnetic field of 0.22 T at an angle of ${60}^o$ with magnetic field. Pitch of helical path traced in $cm$ is [Take mass of proton = $1.6\times {10}^{-27}kg$ and $\pi$ = $\frac{22}{7}$] (Write upto two digits after the decimal point.)

Answer 1

Pitch = $vcos{60}^o\times T$
where, $T=\frac{2\pi m}{qB}$
Pitch$=4\times {10}^5\times \frac{1}{2}\times 2\times \frac{22}{7}\times \frac{1.6\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.22}$
Pitch $=5.71cm$