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 $2 \times 10^{6} \,ms ^{-1}$ at an angle of $60^{\circ}$ to the $X$-axis. If a magnetic field of $0.104 \,T$ is applied along $Y$-axis, then path of proton is

• A. a circle of radius $0.2 \, m$ and time period of $\pi \times 10^{-7} \,s$
• B. a circle of radius $0.1 \, m$ and time period $2 \pi \times 10^{-7} \,s$
• C. a helix of radius $0.2 \,m$ and time period of $2 \pi \times 10^{-7} \, s$
• D. a helix of radius $0.2 \,m$ 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}}{B q}=\frac{1.67 \times 10^{-27} \times 2 \times 10^{6}}{0.104 \times 1.6 \times 10^{-19}} $
$\Rightarrow r=0.2 \,m$
Time period, $T=\frac{2 \pi m}{B q}$
$=2 \pi \times 10^{-7}\, s$