0 votes
in Kinematics by (90.1k points)
edited by
A proton of mass ' \(m\) ' moving with a speed \(v(<<c\), velocity of light in vacuum) completes a circular orbit in time ' \(T\) ' in a uniform magnetic field. If the speed of the proton is increased to \(\sqrt{2} \mathrm{v}\), what will be time needed to complete the circular orbit?
(A) \(\sqrt{2} \mathrm{~T}\)
(B) \(\mathrm{T}\)
(C) \(\frac{T}{\sqrt{2}}\)
(D) \(\frac{\mathrm{T}}{2}\)

3 Answers

0 votes
by (90.1k points)
Ans : (B)  Time period is independent of velocity. So new time period = T
0 votes
by
buy cialis 40mg pill <a href="https://ordergnonline.com/">cialis ca</a> buy ed pills online usa
0 votes
by
overnight delivery for cialis <a href="https://ordergnonline.com/">cialis 5mg pills</a> ed pills comparison

Related questions

...