# Current from magnetic configuration

1. Aug 21, 2008

1. A simple magnetic field configuration is given by $$\vec{B}=ax\hat{y}+by\hat{x}$$. What is the current in this system? Assume that a and b are constant.

2. Maxwells equations.

3. I'm not exactly sure what is being asked here. Off the top of my head I would like to say the answer is $$\vec{E}=ay\hat{y}+bx\hat{x}$$ as B fields are at 90 degrees to E fields, so you simply swap over the X and Y. But this doesn't give you current, rather the E-field.... confused.

Last edited: Aug 21, 2008
2. Aug 21, 2008

### Defennder

The question appears to be oddly stated. Current is flux through some given surface. You could use $$\nabla \times \mathbf{B}$$ from one of Maxwell's equation to give you the current density, but unless the surface is specified, that's all you can ever get from this, unless I'm missing out something here.

3. Aug 21, 2008

### merryjman

I agree with defennder here; doing $$\vec{\nabla} x \vec{B}$$ can get you the current density (and it's a pretty simple answer) but you have to have a defined loop to get the current. But the current density can tell you some important things about the current itself.

I think that the E perpendicular to B that you speak of applies to electromagnetic waves, whose B and E fields have time dependence as well, not necessarily some made-up magnetic field function.

4. Aug 22, 2008

Thanks for the reponse. I'm still a bit confused, maybe it is a bit of an ambiguos question... its not from a textbook but was a written question, so I'm presuming they meant current density...

how exactly would you do $$\vec{\nabla} x \vec{B}$$ to get the current density? not sure what all the symbols mean to be frank. Just started learning this stuff, and some of the maths is beyond me. Once given the answer I can generally work backwards and understand. If only exams were that easy :)

(Would this imply that the current density is 0? or am i misunderstanding this)

5. Aug 22, 2008

### Defennder

$$\nabla \times \mathbf{B} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k} \\ \\ {\frac{\partial}{\partial x}}&{\frac{\partial}{\partial y}}&{\frac{\partial}{\partial z}} \\ \\ B_x&B_y&B_z \end{vmatrix}$$ where the various B(subscripts) are the x,y,z components of B.

6. Aug 22, 2008

Thanks. So.... now I am more confused What exactly do I have to solve from this to get the current density? is it related to Maxwell's equations in terms of E and B for linear materials?

I think what is probably a good idea is someone explaining what all the terms in the original equation represent, $$\vec{B}=ax\hat{y}+by\hat{x}$$. I'm not sure of the difference between the x with a ^ over it or the x without. I thought it was in two dimensions, but you seem to have listed i,j and a 'k', component, which has thrown me a bit.

7. Aug 22, 2008

I checked and it should be current density, not current.

... I think that the current density would be 0. Correct?

8. Aug 22, 2008

### Defennder

The two vertical lines represent the determinant of the 3x3 matrix. That's the definition of curl B. Just find the determinant.

I didn't get zero.

9. Aug 22, 2008

### merryjman

Hmm, I know how it feels to be new at this kind of thing. Defennder gave a good mathematical description of "curl B" in terms of matrices and determinants. Maybe it will help to describe in a hand-wavy fashion? It will be useful to you to draw a few field vectors on the X-Y plane to visualize the field.

The little carat over the x and y tells you that these are vectors. For example, the first part tells you that the field's y-component is ax, and the x-component is by. (like at (1,1), the B vector is <b,a> but at (-2,2) the vector is <2b,-2a>) If you do this, you will see the curly nature pop out.

"curl B" is a good descriptive term because the operation gives you a sense of how curly (or whirlpooly or whatever) the B field is. Static electric field lines tend to "diverge," or spread out, like the field lines from a point charge, so the mathematical operator $$\vec{\nabla}\cdot \vec{E}$$, called the "divergence" operator, sort of tells you how much the electric field spreads out. The dot product gives a scalar which is why the right side of Gauss's Law is just scalars (enclosed charge density & permittivity). Steady-current magnetic fields, on the other hand, don't diverge at all; rather, they loop in circles around the current. The curl operator $$\vec{\nabla}\times \vec{B}$$ tells you how much the B field curls, and in what direction. That's an important aspect of the curl, it gives you a vector rather than a scalar. This should make sense; after all, current (or current density) has to have a direction, but static charge does not.

As to your question about the third dimension k, well, this is a consequence of the 3-D nature of magnetic fields. If you have a wire pointing up, you can call that the k (or z) direction, whereas the direction of the field lines is in the i and j (x and y) directions. Maybe do a google image search for "right hand rule" and "right hand curl rule current" for illustration of this 3-D nature.

So if you do the curl of your particular B field, you will get a vector function which tells you the current density and its direction. With a defined loop, you can figure out how much current is in the loop by integration. As defennder said, if you do the curl correctly you shouldn't get zero. Use your Calc textbook to help with the curl operation.

Last edited: Aug 22, 2008