Question

A proton of mass $1.67\times 10^{-27}kg$ and charge $1.6\times 10^{-19}C$ is projected with a speed of $2\times 10^{6}m{s}^{-1}$ at an angle of $60^{\circ }$ to the $X$-axis. If a magnetic field of $0.104T$ is applied along $Y$-axis, then path of proton is

• A. a circle of radius $0.2m$ and time period of $\pi \times 10^{-7}s$
• B. a circle of radius $0.1m$ and time period $2\pi \times 10^{-7}s$
• C. a helix of radius $0.2m$ and time period of $2\pi \times 10^{-7}s$
• D. a helix of radius $0.2m$ and time period of $4\pi \times 10^{-7}s$

Answer 1

Answer: (C)

As, velocity $v$ component is along magnetic field $B$.
So, path is helical. Radius of a helix,
$r=\frac{m{v}_{\perp }}{Bq}=\frac{1.67\times 10^{-27}\times 2\times 10^6}{0.104\times 1.6\times 10^{-19}}$
$\Rightarrow r=0.2m$
Time period, $T=\frac{2\pi m}{Bq}$
$=2\pi \times 10^{-7}s$