The magnetic field in a certain region of space is given by B =8.35×10⁻² Î T. A proton is shot into the field with velocity v=(2×10⁵i ̂ + 4×10⁵ j ̂ ) m/s. The proton follows a helical path in the field. The distance moved by proton in the x-direction during the period of one revolution in the yz- plane will be (mss of proton=1.67×10⁻²⁷ kg)

Options

(a) 0.053 m
(b) 0.136 m
(c) 0.157 m
(d) 0.236 m

Correct Answer:

0.157 m

Explanation:

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