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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 circle of radius =$ 0.2 \,m$ and time period $\pi \times {10^{ - 7}}\,s$
A circle of radius = $0.1\, m$ and time period $2\pi \times {10^{ - 7}}\,s$
A helix of radius =$0.1 \,m $ and time period $2\pi \times {10^{ - 7}}\,s$
A helix of radius = $0.2\, m$ and time period $4\pi \times {10^{ - 7}}\,s$
Solution
(c) Path of the proton will be a helix of radius $r = \frac{{mv\sin \theta }}{{qB}}$
(where $\theta$ = Angle between $\overrightarrow B \,{\rm{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}}$
$ = 0.1\,m$
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$
