# Movement of Particle in Electromagnetic Field

I'm a high schooler who just finished learning Maxwell's Equations. I'm trying to visualize the movement of a charged particle in a field.

I have an infinite sheet of positive charge oriented on the xz plane and located on the +y axis, moving at a velocity $$v$$ in the $$+x$$ direction. A test charge $$Q$$ of mass $$m$$ also moves at a velocity $$v$$ alongside. The relevant diagram is below.

A stationary observer declares the force on the test charge to be $$F=Q(E+v\times B)$$. Here's where I'm confused: I want to find how $$Q$$ actually moves. I have the magnitude of the Electric & Magnetic Force, but what's the direction? As in, how will the test charge actually move through the field? It is critical for me to understand this, because I'm building a visualization of this exact situation. In my visualization above, we have a discrete array of red particles representative of the Charged Sheet. We further have a test charge (stationary for the time being), whose Net Electric Force is the Red Vector, and Net Magnetic Field is the Cyan Vector. Please advise. Thank you.

Here is an additional diagram that represents the situation I'm interested in visualizing, with the relevant Magnetic and Electric Field Vectors: To calculate the Electric Field, I used Gauss' Law of Electric Fields, as follows, with a Cylinder as my Gaussian Surface (due to the symmetry of the sheet):

\begin{align} \int { E\cdot n }da=\frac { { q }_{ enc } }{ { \varepsilon }_{ 0 } } \\ E(A)=\frac { { q }_{ enc } }{ { \varepsilon }_{ 0 } } \\ E(2\pi { r }^{ 2 })=\frac { \sigma \pi { r }^{ 2 } }{ { \varepsilon }_{ 0 } } \\ E=\frac { \sigma }{ { 2\varepsilon }_{ 0 } } \end{align}

Since both the Test Charge and the Sheet are positively charged, I conjecture the direction to be $$-\hat { j }$$, as they repel each other. And since the Electric Force is proportional to the Electric Field, both the Electric Force and Field are in the $$-\hat { j }$$ direction.

To calculate the Magnetic Field, I used Ampere's Law, using a rectangle as my Amperian Loop as follows:

\begin{align} \int { B\cdot n } ds={ \mu }_{ 0 }{ I }_{ enc }\\ 2Bh={ \mu }_{ 0 }{ I }_{ enc }\\ 2Bh={ \mu }_{ 0 }({ J }_{ s }h)\\ B=\frac { { \mu }_{ 0 }{ J }_{ s } }{ 2 } \end{align}

I conjecture the magnetic field to be in the same direction as the Electric Field such that $$B\perp v$$ so $$B$$ is in the $$-\hat { j }$$ direction. Thus, $$B$$ is in the $$-\hat { j }$$ direction, $${ F }_{ B }$$ is in the $$+\hat { k }$$ direction, and $$v$$ is in the $$+\hat{i}$$ direction.

Main Question: I want to find how $$Q$$ actually moves. I have the magnitude of the Electric & Magnetic Force, but what's the direction?

• oriented on the xz plane and located on the +y axis I can’t figure out what that means. Sep 8 '20 at 0:07
• @G.Smith I've attached an additional diagram now. Please let me know if that's helpful. Sep 8 '20 at 0:07
• Yes, that makes it clear. Sep 8 '20 at 0:12
• You’ve called this question “Movement of Particle in Electromagnetic Field” but it is as much about calculating the magnetic field of a sliding sheet of charge as it is about the motion of the point charge in that field. Sep 8 '20 at 0:14
• You need to show more of your work in calculating that field. Sep 8 '20 at 0:15

This equation you wrote, $$F = Q(E + v \times B)$$, is actually a vector equation: every quantity except $$Q$$ is a vector. In other words it's clearer to write $$\vec{F} = Q(\vec{E} +\vec{v} \times \vec{B})$$.
Now you can see all the directions. If we leave aside the agnetic term for now and only look at $$\vec{F} = Q(\vec{E})$$, then we see that the electric force is in the same direction as the electric field. Similarly, the magnetic term $$\vec{F} = Q\vec{v} \times \vec{B}$$, the magnetic force is in the direction of $$\vec{v} \times \vec{B}$$, and the direction of the cross product is well-defined.
I don't understand your visualization very well but it looks to me like your $$\vec{B}$$ is in an incorrect direction (should be $$-\vec{z}$$ direction, since the current is moving upwards [$$\vec{x}$$ direction]). If we assume that the $$\vec{B}$$ in the diagram is correct, then $$\vec{v} \times \vec{B}$$ would be in the $$-\vec{z}$$ direction, by definition of the cross product.
• To actually find $v$, can I have $v={v}_{0}+at$, where $a=\frac{F}{m}$, and $F=Q(v \times B)$ so that $v=\frac{Q(v \times B)}{m}$. Would that be the way to go? Sep 8 '20 at 1:27
• @DarkRunner not a good idea since in that last equation you have $v$ on both sides of the equation. What exactly is the problem you're trying to solve? Are you given $v$ and asked to find the direction of the magnetic force, or vice versa? Sep 8 '20 at 1:31
• My goal is this: To find the time $t$ the Test Charge $Q$ takes to travel $\Delta y$ according to a Stationary Observer and an Observer on the Sheet of Charge. Sep 8 '20 at 1:44
• @DarkRunner find the force first, then calculate the acceleration from Newton's 2nd law and apply the kinematic equations. In the first comment there actually is a typo, it should be $a = \ldots$ on the left hand side, so it is technically OK (just remember that one $v$ is for the particle and the other $v$ is for the sheet of charge). Sep 8 '20 at 1:50
• Would i.imgur.com/ECd9GRe.png be correct if I had a negative sheet of charge? I have $dir(B)=\hat{k}$ & $dir(E)=\hat{j}$ Sep 8 '20 at 10:58