How do I generate a magnetic vector field using equations?

• darkdave3000
In summary, using a pair of point charges will approximate a magnetic dipole's magnetic field. However, if the charges are close together, the field will be zero. To calculate the vector fields, you need to use the dipole formula or the large-##|\mathbf{x}|## approximation.
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?

darkdave3000 said:
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.

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

darkdave3000 said:
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.

PeroK said:
Static charges give rise to an electric field. To have a magnetic field you need moving charges.
What do you suggest then?

darkdave3000 said:
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?

PeroK said:
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.

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$$\mathbf{B} = \nabla \times \mathbf{A} = \mathbf{e}_i \varepsilon_{ijk} \partial_j A_k$$

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vanhees71
etotheipi said:
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

darkdave3000 said:
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
darkdave3000 said:
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 $$B_x = (\nabla \times \mathbf{A})_x = \mathbf{e}_x \cdot (\nabla \times \mathbf{A})$$
darkdave3000 said:
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

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vanhees71
etotheipi said:
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

darkdave3000 said:
What is the name of this equation?

Kind Regards, David
Which one?

etotheipi said:
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

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.$$\mathbf{e}_i \cdot \sum_j u_j \mathbf{e}_j = \sum_j u_j \delta_{ij} = u_i$$

hutchphd and vanhees71
darkdave3000 said:
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

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.

vanhees71
hutchphd said:
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

1. How do I calculate the magnitude of a magnetic vector field?

The magnitude of a magnetic vector field can be calculated using the equation B = μ0I/2πr, where B is the magnetic field strength, μ0 is the permeability of free space, I is the current, and r is the distance from the current.

2. What is the equation for determining the direction of a magnetic vector field?

The direction of a magnetic vector field can be determined using the right-hand rule. Point your thumb in the direction of the current, and your fingers will curl in the direction of the magnetic field.

3. How do I generate a uniform magnetic vector field?

To generate a uniform magnetic vector field, you can use the equation B = μ0NI/L, where B is the magnetic field strength, μ0 is the permeability of free space, N is the number of turns in the coil, and L is the length of the coil.

4. What is the difference between a magnetic vector field and an electric vector field?

A magnetic vector field is created by moving electric charges, while an electric vector field is created by stationary electric charges. Additionally, the direction of a magnetic vector field is perpendicular to the direction of the current, while the direction of an electric vector field is parallel to the direction of the electric field.

5. How can I use equations to manipulate a magnetic vector field?

There are several equations that can be used to manipulate a magnetic vector field, such as the superposition principle, which states that the total magnetic field at a point is the vector sum of the individual magnetic fields at that point. Additionally, the magnetic field can be manipulated by changing the current, the distance from the current, or the number of turns in a coil.

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