A proton of mass 1.6×10-27 kg and charge 1.6×10-19 C is projected with a speed of 2×106m/s at an angle of 60° to the x-axis. If a uniform magnetic field of 0.104 tesla is applied along y-axis, the path of proton is

r = mvsin30° = (1.67×10-27)×(2×106)sin30°1.6×10-19×0.104 = 0.1m T =2πmqB = 2π×1.67×10-271.6×10-19×0.104 = 2π×10-7s

A proton of mass 1.6×10-27 kg and charge 1.6×10-19 C is projected with a speed of 2×106m/s at an angle of 60° to the x-axis. If a uniform magnetic field of 0.104 tesla is applied along y-axis, the path of proton is

r = mvsin30° = (1.67×10-27)×(2×106)sin30°1.6×10-19×0.104 = 0.1m T =2πmqB = 2π×1.67×10-271.6×10-19×0.104 = 2π×10-7s

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A proton of mass $1.6 \times 10^{- 27} k g$ and charge $1.6 \times 10^{- 19} C$ is projected with a speed of $2 \times 10^{6} m / s$ at an angle of $60 °$ to the $x -$ axis. If a uniform magnetic field of $0.104 t e s l a$ is applied along $y -$ axis, the path of proton is

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a circle of radius $= 0.2 m$ and time period $π \times 10^{- 7} s$

a circle of radius $= 0.1 m$ and time period $2 π \times 10^{- 7} s$

a helox of radius $= 0.1 m$ and time period $4 π \times 10^{- 7} s$

a helox of radius $= 0.2 m$ and time period $2 π \times 10^{- 7} s$

answer is C.

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Detailed Solution

$r = m v \sin 30 ° = \frac{(1.67 \times 10^{- 27}) \times (2 \times 10^{6}) \sin 30 °}{1.6 \times 10^{- 19} \times 0.104} = 0.1 m T = \frac{2 π m}{q B} = \frac{2 π \times 1.67 \times 10^{- 27}}{1.6 \times 10^{- 19} \times 0.104} = 2 π \times 10^{- 7} s$