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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
- The concentric spheres of radii R and r have positive charges q₁ and q₂ with equal
- A pure semiconductor behaves slightly as a conductor at
- Which of the following transitions will have highest emission wavelength?
- 6Ω and 12Ω resistors are connected in parallel. This combination is connected
- What is the value of Ā+A in the Boolean algebra?
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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