How do I generate a magnetic vector field using equations?

[darkdave3000]

I am considering using a pair of point charges: positive and negative electric charge to model a magnetic dipole's magnetic field by just average the electric field vectors between the two charged particles where they overlap. Will that work?

In this case the + field will be vectors pointing away from the origin of the positively charged particle(positive divergence).
The - particle will have vectors all pointing toward it according to the electric field strength related to distance of a point particle same as above but backwards.

Then I just average all the arrows between where they overlap?

Or do I have to use the formula here:
https://en.wikipedia.org/wiki/Magnetic_vector_potential

To calculate the X,Y and Z components of each magnetic field vector?

 

[hutchphd]

Summary:: How do I calculate each vector's magnitude and direction in a vector field representing magnetic field of a magnetic dipole given some initial values?

The - particle will have vectors all pointing toward it according to the electric field strength related to distance of a point particle same as above but backwards.

The point is that it is not quite exactly the same. If it were exactly the same the result would be zero everywhere.
If one is far away (compared to the separation) one can use Taylor expansion to generate an accurate approximation: this is the dipole formula.

 

[darkdave3000]

I know that if the positive and negative particle are sharing the same spot the vector fields of both particles will cancel out. But I am assuming there is some distance between them. So let's say I want to simulate the Earth's magnetic field, can I then separate them at 6000km and then give each particle a very large electric charge each to generate an electric field for each and then average both fields together to get my magnetic vector field?

David

 

[PeroK]

I know that if the positive and negative particle are sharing the same spot the vector fields of both particles will cancel out. But I am assuming there is some distance between them. So let's say I want to simulate the Earth's magnetic field, can I then separate them at 6000km and then give each particle a very large electric charge each to generate an electric field for each and then average both fields together to get my magnetic vector field?

David

Static charges give rise to an electric field. To have a magnetic field you need moving charges.

 

[darkdave3000]

Static charges give rise to an electric field. To have a magnetic field you need moving charges.

What do you suggest then?

 

[PeroK]

What do you suggest then?

I don't think I understand what you want to do. Do you want the field of a pure magnetic dipole?

 

[darkdave3000]

I don't think I understand what you want to do. Do you want the field of a pure magnetic dipole?

YES! I need a formula to calculate the vector's X, Y and Z component with magnitude and direction at each point surrounding the dipole.

Based upon initial condition of the cartesian coordinates of location of the dipole, it's magnetic strength and it's direction given by a vector.

 

[etotheipi]

You can do that like how @hutchphd mentioned in the large-$|\mathbf{x}|$ approximation, i.e. from starting with the vector potential of a current loop containing current $I$,$$\mathbf{A}(\mathbf{x}) = \frac{\mu_0 I}{4\pi} \oint_{\Gamma} \frac{d\mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|}$$about the curve $\Gamma$ along the ring. Now Taylor expand the integrand, noting that$$\frac{1}{|\mathbf{x} - \mathbf{x}'|} = \frac{1}{|\mathbf{x}|} - \mathbf{x}' \cdot \nabla \left( \frac{1}{|\mathbf{x}|} \right) + \dots = \frac{1}{|\mathbf{x}|} + \mathbf{x}' \cdot \frac{1}{|\mathbf{x}|^2} \mathbf{e}_r + \dots = \frac{1}{|\mathbf{x}|} + \frac{\mathbf{x}' \cdot \mathbf{x}}{|\mathbf{x}|^3} + \dots $$where $\mathbf{e}_r = \mathbf{x}/ \mathbf{|x|}$. Then use the result that $$\oint_{\partial S} d\mathbf{x}' (\mathbf{x} \cdot \mathbf{x}') = \left( \int_S d\mathbf{S} \right) \times \mathbf{x}$$ where $\tilde{\mathbf{S}} = \left( \int_S d\mathbf{S} \right)$ is the vector area bounded by the $\partial S$, to simplif your answer. You can also use the definition $\mathbf{m} = I \tilde{\mathbf{S}}$ of the magnetic moment. Then all you need to do is take the curl$$\mathbf{B} = \nabla \times \mathbf{A} = \mathbf{e}_i \varepsilon_{ijk} \partial_j A_k $$

 

[darkdave3000]

You can do that from starting with the vector potential of a current loop containing current $I$,$$\mathbf{A}(\mathbf{x}) = \frac{\mu_0 I}{4\pi} \oint_{\Gamma} \frac{d\mathbf{x}'}{|\mathbf{x} - \mathbf{x}'|}$$about the curve $\Gamma$ along the ring. Now Taylor expand the integrand, noting that$$\frac{1}{|\mathbf{x} - \mathbf{x}'|} = \frac{1}{|\mathbf{x}|} - \mathbf{x}' \cdot \nabla \left( \frac{1}{|\mathbf{x}|} \right) + \dots = \frac{1}{|\mathbf{x}|} + \mathbf{x}' \cdot \frac{1}{|\mathbf{x}|^2} \mathbf{e}_r + \dots = \frac{1}{|\mathbf{x}|} + \frac{\mathbf{x}' \cdot \mathbf{x}}{|\mathbf{x}|^3} + \dots $$where $\mathbf{e}_r = \mathbf{x}/ \mathbf{|x|}$. Then use the result that $$\oint_{\partial S} d\mathbf{x}' (\mathbf{x} \cdot \mathbf{x}') = \left( \int_S d\mathbf{S} \right) \times \mathbf{x}$$ where $\tilde{\mathbf{S}} = \left( \int_S d\mathbf{S} \right)$ is the vector area bounded by the $\partial S$, to simplif your answer. You can also use the definition $\mathbf{m} = I \tilde{\mathbf{S}}$ of the magnetic moment. Then all you need to do is take the curl$$\mathbf{B} = \nabla \times \mathbf{A} = \mathbf{e}_i \varepsilon_{ijk} \partial_j A_k $$

Thanks for this! I need to catch up with to your level of physics. So this equation, what is it called? The Taylor Expression?

The "x" is that the x component of the magnetic field at a set distance and angle relative to the original magnetic dipole?

Is there a wikipedia article that can help me understand this formula and the variables in it?

David

David

 

[etotheipi]

So this equation, what is it called? The Taylor Expression?

For a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ like this you have$$f(\mathbf{x} + \delta \mathbf{x}) = f(\mathbf{x}) + \delta \mathbf{x} \cdot \nabla f(\mathbf{x}) + \frac{1}{2}(\delta \mathbf{x} \cdot \nabla)^2 f(\mathbf{x}) + \dots$$that is the Taylor expansion

The "x" is that the x component of the magnetic field at a set distance and angle relative to the original magnetic dipole?

No, $\mathbf{x}$ is the position vector of the point at which you are evaluating the vector potential, $\mathbf{x}'$ is the integration variable over the boundary of the curve comprising the dipole, and $\mathbf{A}$ is the vector potential.

The magnetic field at $\mathbf{x}$ would be $\mathbf{B}= \nabla \times \mathbf{A}$, and the $x$ component of the magnetic field would be $$B_x = (\nabla \times \mathbf{A})_x = \mathbf{e}_x \cdot (\nabla \times \mathbf{A})$$

Is there a wikipedia article that can help me understand this formula and the variables in it?

All relevant theory here is covered in Griffiths' EM book

 

[darkdave3000]

For a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ like this you have$$f(\mathbf{x} + \delta \mathbf{x}) = f(\mathbf{x}) + \delta \mathbf{x} \cdot \nabla f(\mathbf{x}) + (\delta \mathbf{x} \cdot \nabla)^2 f(\mathbf{x}) + \dots$$that is the Taylor expansionNo, $\mathbf{x}$ is the position vector of the point at which you are evaluating the vector potential, $\mathbf{x}'$ is the integration variable over the boundary of the curve comprising the dipole, and $\mathbf{A}$ is the vector potential.

The magnetic field at $\mathbf{x}$ would be $\mathbf{B}= \nabla \times \mathbf{A}$, and the $x$ component of the magnetic field would be $$B_x = (\nabla \times \mathbf{A})_x = \mathbf{e}_x \cdot (\nabla \times \mathbf{A})$$All relevant theory here is covered in Griffiths' EM book

What is the name of this equation?

Kind Regards, David

 

[etotheipi]

What is the name of this equation?

Kind Regards, David

Which one?

 

[darkdave3000]

The magnetic field at $\mathbf{x}$ would be $\mathbf{B}= \nabla \times \mathbf{A}$, and the $x$ component of the magnetic field would be $$B_x = (\nabla \times \mathbf{A})_x = \mathbf{e}_x \cdot (\nabla \times \mathbf{A})$$All relevant theory here is covered in Griffiths' EM book

This part

 

[etotheipi]

Let a vector field $\mathbf{v} : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ have zero divergence, $\nabla \cdot \mathbf{v} = 0$. Then, $\mathbf{v}$ can be expressed as the curl of another vector field $\mathbf{W}$, i.e.$$\mathbf{v} = \nabla \times \mathbf{W}$$by Helmholtz' theorem. Maxwell's second equation says that $\mathbf{B}$ is divergence free, and consequently we can express $\mathbf{B} = \nabla \times \mathbf{A}$. Here $\mathbf{A}$ is called the vector potential. It is not unique, and only defined up to addition of the gradient of a function, i.e. $\mathbf{A}' = \mathbf{A} + \nabla f$ will also give $\nabla \times \mathbf{A}' = \nabla \times \mathbf{A} = \mathbf{B}$.

To take the $i^{\text{th}}$ component of a vector $\mathbf{u} = \sum_j u_j \mathbf{e}_j$, so long as your coordinates are orthonormal (i.e. $\mathbf{e}_a \cdot \mathbf{e}_b = \delta_{ab}$), you can just take scalar product of the basis vector with the vector, i.e.$$\mathbf{e}_i \cdot \sum_j u_j \mathbf{e}_j = \sum_j u_j \delta_{ij} = u_i$$

 

[PeroK]

YES! I need a formula to calculate the vector's X, Y and Z component with magnitude and direction at each point surrounding the dipole.

Based upon initial condition of the cartesian coordinates of location of the dipole, it's magnetic strength and it's direction given by a vector.

Try this, and/or search for more on "magnetic dipole":

http://www.cbpf.br/~ieeemag/lectures/Jander_I.pdf

 

[hutchphd]

I think perhaps I misunderstood the original question. If you have two magnetic dipoles the resultant vector B field is the vector sum of the vector field from each individually. For the individual dipoles see for instance:
https://en.wikipedia.org/wiki/Magnetic_dipole
I thought you were trying to derive this result.

 

[darkdave3000]

I think perhaps I misunderstood the original question. If you have two magnetic dipoles the resultant vector B field is the vector sum of the vector field from each individually. For the individual dipoles see for instance:
https://en.wikipedia.org/wiki/Magnetic_dipole
I thought you were trying to derive this result.

So for example say I want to model the Earth's magnetic field, can I just sum the vector fields of two poles 12,000km apart? Will these poles be oppositely charged electric particles or magnetic poles?

And which formula would I use to generate the field for each pole before "summing them"?

David

 

[darkdave3000]

Try this, and/or search for more on "magnetic dipole":

http://www.cbpf.br/~ieeemag/lectures/Jander_I.pdf

Thanks for this theoretical material! I will study!