# What affects the particle first?

I just want to check my reasoning. We have a particle starting at the origin with

$$v(0)=\frac{E}{B}\hat y$$

There is a magnetic field pointing in the $\hat x$ direction and an electric field pointing in the $\hat z$ direction. What happens right after $t=0$?

I think one of these 3 things will happen:

1. If the magnetic field affects the particle first, then it will initially start to curve downwards.

2. If the E-field affects the particle first, then it will initially curve upwards.

3. If both fields affect the particle initially, then I think the particle will move in a straight line from the origin along the $y$-axis.

• I added an explanation if my previous answer was kinda superficial. – Prasad Mani Oct 24 '16 at 3:48

If both the electric and the magnetic field are present at the start and the given velocity in y-direction is $v(0)= E/B$, then the magnetic Lorentz force component $q\vec v×\vec B$ directed in negative z-direction has the same strength as the electric field force $q\vec E$ directed in positive z-direction. Thus the magnetic and electric field forces compensate each other and the particle will fly with constant velocity $v=v(0)$ along the y-axis.

• freecharly, please see my comment to Prasad Mani. What I'm missing? – HolgerFiedler Oct 24 '16 at 17:54
• @HolgerFiedler - I have looked at you comment and think that it is not correct. In case 3, from the beginning, the total force in on the particle is the Lorentz force $\vec F=q\vec E + q\vec v×\vec B$. Here it is assumed that the velocity $v(0)=E/B$ is in y-direction so that with $q>0$ the magnetic force is $q(E/B)·B=qE$ is in negative z-direction while the electric field force $qE$ is positive z-direction. The electric and magnetic field forces cancel each other exactly giving a zero net force. Thus the particle continues its flight on a straight line with constant velocity $v=v(0)$. – freecharly Oct 24 '16 at 18:59
• @HolgerFiedler - The situation is similar to case of the stationary Hall effect. There after the build up of the electric field counteracting the magnetic field force, the charge carriers move on straight path. – freecharly Oct 24 '16 at 20:00

Option 1 and 2 are only valid when one of the fields is turned on before the other so that it affects the charged particle first. Otherwise, option 3 is the correct one. Particle travels with the same velocity, undeflected

## Explanation

The thing is we just have to check for the force acting on the particle at any instant. The electric field will exert a force $F_E$ = $qE\hat{z}$ in the positive $z$ direction

The magnetic field will exert a force $F_B$ given by -

$F_B$ = $q\vec{v}\times\vec{B}$ = $q\frac{E}{B}\hat{y}\times B\hat{x}$ = $-qE\hat{z}$

This, as you can see is time independent, meaning no matter which instant of time you look at, you will see $F_E$ = $-F_B$.....cancelling each other perfectly. So the particle will move unfettered. Had there been an imbalance of force at any point of time, we would have seen cycloidal motion. But not here!

• PrasadMani, I'm not sure about the straight line in case 3. Even if in a starting state both fields will be of the same influence to the charged particle the magnetic field will brake down the electron and move it in a spiral path meanwhile the electric field will accelerate the electron sideways. The first is a trigonometric function while the second is a quadratic function. In result this could not be a straight line. Or do I something missing? – HolgerFiedler Oct 23 '16 at 20:17