Magnetic Fields and Forces

[soul5]

Homework Statement

Prove that the time for one revolution of a charged particle moving perpendicular to a uniform magnetic field is independent of it's speed. Write an expression that gives the period T in terms of the mass and charge of the particle and the magnetic field strength.

Homework Equations

Fc=4*pi*^2mr/T^2

Fm=qvB

The Attempt at a Solution

Fc=Fm

v=4*pi*^2mr/T^2qB

is that right?

 

[alphysicist]

Hi soul5,

You're expressions look like they are true, but they don't answer the question. The question asks for the expression for the period, but that expresion is not supposed to have r and v in it.

If you can find a formula for the period in terms of r and v you can use that to eliminate r and v in your equation.

So let's say you have an object moving at speed v in a circular path of radius r. What is the period (time to go once around the circle)? Once you have that, use it to eliminate r and v. What do you get?

 

[Moetasim]

This is not entirely correct. The correct equation for the period of a charged particle moving perpendicular to a uniform magnetic field is T = 2*pi*m/qB, where m is the mass of the particle, q is its charge, and B is the magnetic field strength. This equation can be derived by setting the magnetic force (Fm = qvB) equal to the centripetal force (Fc = mv^2/r) and solving for T. This shows that the period is independent of the particle's speed, as the velocity term cancels out. This also makes intuitive sense since the magnetic field only affects the direction of the particle's motion, not its speed. Additionally, this equation is consistent with the units of time (seconds) and the fact that the period should decrease as the magnetic field strength increases.