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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:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A geostationary satellite is orbiting the earth at a height of 5R above that surface
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Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A geostationary satellite is orbiting the earth at a height of 5R above that surface
- A body is under stress F and has Young’s modulus Y, then the energy
- The S.I. unit of electric flux is
- Which is correct relation?
- A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV.
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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