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  • A proton of mass  and charge <img alt="1.67 \times 10^{-19}C"

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

Option: 1

A circle of radius=0.2m and a time period \pi \times 10^{-7}


Option: 2

A circle of radius=0.1m and a time period 2\pi \times 10^{-7}


Option: 3

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


Option: 4

A helix of radius=0.2m and time period 4\pi \times 10^{-7}


Answers (1)

best_answer

The radius of the helical path - r=\frac{m(v\sin \theta )}{qB}

 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

Time period T=\frac{2\pi m}{qB}= \frac{2\pi \times 1.67\times10^{-27}}{1.6\times10^{-19}\times0.104}= 2\pi \times10^{-7}sec

 

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Nehul

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