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 $$\vec{v}$$ in a magnetic field $$\vec{B}$$ is equal to $$q\vec{v}\times\vec{B}$$. If a charged particle moves in a circular orbit with a fixed speed $$v$$ 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

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

Homework Equations

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

The Attempt at a Solution

The particle moves in a circle then the magnetic field is perpendicular to the velocity and $$F=qvB$$.

$$f=\frac{v}{2\pi R}$$

$$qvB=\frac{ma}{(1-v^2/c^2)^{3/2}}<br /> =\frac{m}{(1-v^2/c^2)^{3/2}}\frac{v^2}{R}$$

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

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

What's wrong?

 

[Maxim Zh]

The relativistic form of Newton's second law is

$$\frac{\partial\vec{p}}{\partial t} = \vec{F},$$

where

$$\vec{p} = m\gamma(v)\vec{v}.$$

The factor

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

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

Good luck!

 

[andresordonez]

Thanks!