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 at an angle of ${60}^{\circ }$ to the X-axis. If a uniform magnetic field of 0.104 Tesla is applied along Y-axis, the path of proton is

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

Answer 1

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

By using $r=\frac{mvsin\theta }{qB}\Rightarrow r=\frac{1.67\times {10}^{-27}\times 2\times {10}^6\times sin{30}^{\circ }}{1.6\times {10}^{-19}\times 0.104}=0.1m$
and it's time period $T=\frac{2\pi m}{qB}=\frac{2\times \pi \times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}\times 0.104}=2\pi \times {10}^{-7}sec$.