# A Plot of Magnetic Field Lines

1. Oct 12, 2012

### KitchiRUs

Hi,

I was wondering - In the plots that we see of magnetic lines of force (like this one) what exactly are they plotting against what?

Meaning is it Bx vs x vs y, or is it Bx vs By vs Bz?

2. Oct 12, 2012

### Khashishi

It plots the trajectory formed by following the magnetic field. The magnetic field line of force is not a vector field. It is a tangent curve to the vector field B.

If you were a "magnetic charge", pushed around by the magnetic field, you would trace out the path of the magnetic field line.

If you think of the magnetic field as an arrow at each point, then the line of force is what you get by connecting all these arrows together into a continuous trajectory.

3. Oct 13, 2012

### KitchiRUs

Okay, right. That makes sense.
I'm attempting to reproduce those plots, in 2D and in 3D, but I still don't fully understand what it means when I plot Bx vs By or Bx vs x vs y (in 3D).
Would Bx vs x vs y represent the gradient of the field in the xy plane?

4. Oct 15, 2012

### Khashishi

$B_x$ means the component of B along the x direction. $\mathbf{B}$ is a vector field, which means there is a vector $\mathbf{B}(x,y,z)$ at every point in space (ignoring the time coordinate). So there are three scalar functions of three variables, $B_x(x,y,z), B_y(x,y,z), B_z(x,y,z)$. Vectors in 3D can be specified using 3 scalars.

It doesn't really make sense to plot $B_x(x,y,z) vs B_y(x,y,z)$ without specifying what x, y, and z are.

Gradient is something different. It is the slope of a scalar field. The gradient of $B_x$ is a vector field, while the gradient of $\mathbf{B}$ is a rank 2 tensor because $\mathbf{B}$ is already a vector field

5. Oct 15, 2012

### mikeph

You're not plotting B at all, you're plotting a family of parametric curves which are at every point tangential to B.

6. Oct 16, 2012

### KitchiRUs

I've seen those differential equations for $\textbf{B}$. Except in 3D they are three coupled ODE's no? Is there an easier way to solve for the tangent at each point rather than to solve the ODE equations?

Also, I plotted $\textsl{B}_{x}$ vs $\textsl{B}_{y}$, and it looked like the dipole field plots that we see in textbooks... After reading what you guys have said I'm confused about why that happens.

7. Oct 16, 2012

### mikeph

The equation for one of your streamlines will be x(t), where t increases from 0 and x(t) is the streamline curve equation. If you write the component forms, ie.

x(t) = [x(t), y(t), z(t)]
B(x) = [Bx(x,y,z), By(x,y,z), Bz(x,y,z)]

dx(t)/Bx = dy(t)/By = dz(t)/Bz,

Which can be cross-multiplied to give your equations (dx/dy = Bx/By, dy/dz = By/Bz, dz/dx = Bz/Bx), then integrated to find x(t) at different "boundary" conditions values of x(0). I think that when you plotted Bx v By you were plotting the slope field of the streamline function in two dimensions, since Bx/By = dx/dy. That might explain the physical significance of what you saw.

The difficulty of solving these depends on the form of B, and in general you will need a solver algorithm.

8. Oct 16, 2012

### KitchiRUs

@MikeyW - Thanks! That really cleared it up for me.

I was looking at the differential equations I had... I just didn't make the connection between them and what I was plotting.