| ⇦ | 
| ⇨ |
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: - Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV.
- An electron moves on a straight line path XY as shown. The abcd is a coil adjacent
- A disc of mass 100 g is kept floating horizontally in air by firing bullets
- What is the energy released by fission of 1 g of U²³⁵?
- A pure semiconductor behaves slightly as a conductor at
Topics: Magnetic Effects of Current and Magnetism
(167)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV.
- An electron moves on a straight line path XY as shown. The abcd is a coil adjacent
- A disc of mass 100 g is kept floating horizontally in air by firing bullets
- What is the energy released by fission of 1 g of U²³⁵?
- A pure semiconductor behaves slightly as a conductor at
Topics: Magnetic Effects of Current and Magnetism (167)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply