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 $0.1 \,T$ 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

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