Question

A proton is projected with a speed of $2\times 10^{6}m/s$ at an angle $60^{\circ }$ to $x$-axis. If a uniform magnetic field of $0.1T$ is applied along $y$-axis, then the path of proton is helical with time period $a\times 10^{-7}s$. Find $a$. (Take $\pi =3.14$ )

Answer 1

Answer: 6.28

For acute angle projection in uniform magnetic field, the charge will moves along helical path. So, radius of helix is
$r=\frac{mv\sin \theta }{qB}$, where $\theta =30^{\circ }$
Time period of helix is
$T=\frac{2\pi m}{qB}$