Relativistic form of Newton's second law

[andresordonez]

SOLVED
(Problem 10, Chapter 2, Modern Physics - Serway)

Homework Statement

Recall that the magnetic force on a charge q moving with velocity [tex]\vec{v}[/tex] in a magnetic field [tex]\vec{B}[/tex] is equal to [tex]q\vec{v}\times\vec{B}[/tex]. If a charged particle moves in a circular orbit with a fixed speed [tex]v[/tex] in the presence of a constant magnetic field, use the relativistic form of Newton's second law to show that the frequency of its orbital motion is

[tex]
f=\frac{qB}{2\pi m}(1-\frac{v^2}{c^2})^{1/2}
[/tex]

Homework Equations

[tex]
F=\frac{ma}{(1-v^2/c^2)^{3/2}}
[/tex]

The Attempt at a Solution

The particle moves in a circle then the magnetic field is perpendicular to the velocity and [tex]F=qvB[/tex].

[tex]
f=\frac{v}{2\pi R}
[/tex]

[tex]
qvB=\frac{ma}{(1-v^2/c^2)^{3/2}}
=\frac{m}{(1-v^2/c^2)^{3/2}}\frac{v^2}{R}
[/tex]

[tex]
R=\frac{mv}{(1-v^2/c^2)^{3/2}qB}
[/tex]

[tex]
f=\frac{(1-v^2/c^2)^{3/2}qB}{2\pi m}
[/tex]

What's wrong?

 

[Maxim Zh]

The relativistic form of Newton's second law is

[tex]
\frac{\partial\vec{p}}{\partial t} = \vec{F},
[/tex]

where

[tex]
\vec{p} = m\gamma(v)\vec{v}.
[/tex]

The factor

[tex]
\gamma(v) = \frac{1}{\sqrt{1-(v/c)^2}}
[/tex]

is constant in this task. Do not differentiate it!
Write the differential equation system for [tex]v_x[/tex] and [tex]v_y[/tex] and derive the frequency.

Good luck!

 

[andresordonez]

Thanks!