# Electron in a Magnetic Field

• jegues
In summary, an electron with a speed of 10^6 m/s enters a 6mT uniform magnetic field that is perpendicular to its velocity. To calculate the time point at which the electron will move in the opposite direction from its final velocity, we use the equations for the force and acceleration of a charged particle in a magnetic field. However, since the magnetic force does not change the speed of the electron, it will move in a circular path with a radius determined by the mass, charge, velocity, and magnetic field strength. It is important to keep track of the direction of the force and to visualize the path of the electron in order to correctly solve the problem.

## Homework Statement

An electron with speed of $$1o^{6} m/s$$ enters a $$6mT$$
uniform magnetic field that is perpendicular to the velocity of the electron. What is one possible time point after the electron entered the magnetic field that the electron will move along the opposite direction from its final velocity?

## The Attempt at a Solution

Since the velocity of the electron and the magnetic field are perpendicular,

$$F_{B} = qv_{o}B = ma$$

So,

$$a = \frac{qv_{o}B}{m}$$

But,

$$v_{f} = 0 = v_{o} + at$$

So,

$$t = \frac{-m}{qB}$$

Where am I going wrong?

It's extremely important to keep track of the direction of the force. Since the magnetic force is always perpendicular to the velocity, it does not change the speed of the electron. You should try to visualize the path of the electron and rethink the question.

It should pull the electron into a circular path, correct ?

Bump, still trying to figure this one out, can I get another hint?

Yes, it will be a circular path. You know the speed of the electron and can calculate the force. Can you figure out the radius of the orbit?

fzero said:
Yes, it will be a circular path. You know the speed of the electron and can calculate the force. Can you figure out the radius of the orbit?

$$F_{B} = qv \times B$$

So,

$$qv \times B = \frac{mv^{2}}{r}$$

$$r = \frac{mv^{2}}{qvBcos\theta}$$

Is theta changing though? Because as the force starts to push the electron down into a circular path the angle between v and B must be different, correct?

You're told that $$\vec{v}$$ and $$\vec{B}$$ are perpendicular, so you might want to rethink how you computed the cross product. You should probably draw a diagram to see the relative directions of $$\vec{v}$$, $$\vec{B}$$ and $$\vec{F}$$.

fzero said:
You're told that $$\vec{v}$$ and $$\vec{B}$$ are perpendicular, so you might want to rethink how you computed the cross product. You should probably draw a diagram to see the relative directions of $$\vec{v}$$, $$\vec{B}$$ and $$\vec{F}$$.

Whoops, I thought they would no longer be perpendicular once the electron started moving in a circular path but that's not true.

So, $$\theta = 0^{o}$$

Therefore,

$$r = \frac{mv^{2}}{qvB} = \frac{mv}{qB}$$