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 ${60}^0$ to the X-axis.If a uniform magnetic filed of 0.104 tesla is applied along Y-axis, the path of proton is

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

Answer 1

Answer: (C)

A helix of radius$=0.1m$ and time period =$2\pi \times {10}^{-7}s$

Path of the proton will be a helix of radius $r=\frac{mv\sin \theta }{qB}$

(where $\theta =$ Angle between $\overrightarrow{B}$ and $\overrightarrow{v}$)

$\Rightarrow r=\frac{1.67\times {10}^{-27}\times 2\times {10}^6\times \sin {30}^o}{1.6\times {10}^{-19}\times 0.104}$

$=0.1m$

Time period $T=\frac{2\pi m}{qB}=\frac{2\pi \times 1.67\times {10}^{-27}}{1.6\times {10}^{-19}\times 0.104}$

$=2\pi \times {10}^{-7}sec$