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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.
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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
- Which of the following particles will have minimum frequency of revolution
- A rocket with a lift-off mass 3.5 x 10⁴ kg is blasted upwards with an initial acceleration
- A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force
- The kirchoff’s first law (∑i=0) and second law (∑iR=∑E), where the symbols
- In communication with help of antenna if height is doubled then the range
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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