The magnetic field in a certain region of space is given by B =8.35×10⁻² Î T.

⇦

⇨

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:

1. A proton of mass 1.6×10⁻²⁷ kg goes round in a circular orbit of radius 0.10 m under
2. The maximum and minimum intensities of two sources is 4:1. The ratio of amplitude is
3. A short circuited coil is placed in a magnetic field varying with time. Electrical
4. A 1 µA beam of protons with a cross-sectional area of 0.5 sq.mm is moving with a velocity
5. The wavelength of light emitted from second orbit to first orbit in a hydrogen atom is

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