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A basic question about electromagnetism


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Hi everybody, I was just curious to know why is it that an electric current; the magnetic field it produces, as well as the direction of the force it produces are each at right angles to each other? So far I have just learned this as a fact with things like Fleming's left hand rule but I was curious to know why such a relationship exists.

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They come about due to Maxwell's Equations. These are 4 equations dealing with how electric and magnetic fields behave. Together, they bring about the rules that those vectors you mentioed are right angles to each other.

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They come about due to Maxwell's Equations[/b']. These are 4 equations dealing with how electric and magnetic fields behave. Together, they bring about the rules that those vectors you mentioed are right angles to each other.

 

So prior to Maxwell the Universe didn't behave this way?

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They come about due to Maxwell's Equations. These are 4 equations dealing with how electric and magnetic fields behave. Together, they bring about the rules that those vectors you mentioed are right angles to each other.

Perhaps you could show HOW this comes out of the Maxwell eqs for a start?

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Allow me.

 

The Maxwell Equations are:

 

[math] \nabla \cdot \mathbf{E} = 4 \pi \rho \qquad \qquad \nabla \times \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}[/math]

 

[math] \nabla \cdot \mathbf{B} = 0 \qquad \qquad \nabla \times \mathbf{B} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}+ \frac{4\pi}{c} \mathbf{J}[/math]

 

where [math]\mathbf{E}[/math] and [math]\mathbf{B}[/math] are the electric and magnetic fields, [math]\mathbf{J}[/math] is the current and [math]\rho[/math] is the charge density.

 

For simplicity lets imagine that the Electric field is constant with time, so that

 

[math] \nabla \times \mathbf{B} = \frac{4\pi}{c} \mathbf{J}[/math]

 

Now, remember that the cross-product of two vectors is at right angles to both of the vectors, so [math] \mathbf{B}[/math] must be at right angles to [math] \nabla \times \mathbf{B}[/math] and therfore must also be at right angles to [math]\mathbf{J}[/math].

 

In other words, the magnetic field is at right angles to the current.

 

A more interesting question is where do these laws come from....?

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I see three fundamental questions, which never seem to appear in other people's ten most important Q list. The first is just above. The second is of the nature of the not-vacuum, which I am dealing with in my thesis. The third is the WTF of gravitation, which is sourced and experienced by all forms. Quantum non-locality will make a fool out of you also, but at least it's somehow connected to #2.....................Planck's constant is of angular momentum and bespeaks cranking, literally. I say that angular momentum density of a field is A x (Adot+del(U) - ®rho) and in bound states the cranking 'meets its own tail' and must be quantized. Am I saying WHAT IT IS? Nope.

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Allow me.

 

The Maxwell Equations are:

 

[math] \nabla \cdot \mathbf{E} = 4 \pi \rho \qquad \qquad \nabla \times \mathbf{E} = -\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}[/math]

 

[math] \nabla \cdot \mathbf{B} = 0 \qquad \qquad \nabla \times \mathbf{B} = \frac{1}{c} \frac{\partial \mathbf{E}}{\partial t}+ \frac{4\pi}{c} \mathbf{J}[/math]

 

where [math]\mathbf{E}[/math] and [math]\mathbf{B}[/math] are the electric and magnetic fields' date=' [math']\mathbf{J}[/math] is the current and [math]\rho[/math] is the charge density.

 

For simplicity lets imagine that the Electric field is constant with time, so that

 

[math] \nabla \times \mathbf{B} = \frac{4\pi}{c} \mathbf{J}[/math]

 

Now, remember that the cross-product of two vectors is at right angles to both of the vectors, so [math] \mathbf{B}[/math] must be at right angles to [math] \nabla \times \mathbf{B}[/math] and therfore must also be at right angles to [math]\mathbf{J}[/math].

 

In other words, the magnetic field is at right angles to the current.

 

A more interesting question is where do these laws come from....?

what is the [math]\nabla[/math] vector?

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No. [math] \vec \nabla = \left( \begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{array} \right) [/math], in the case of cartesian coordinates.

 

Therefore, [math]\vec \nabla \cdot \vec E = \left( \frac{\partial}{\partial x} \ \frac{\partial}{\partial y} \ \frac{\partial}{\partial z} \right) \left( \begin{array}{c} E_x \\ E_y \\ E_z \end{array} \right) = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} [/math]

 

x-hat, y-hat and z-hat were supposed to be the base vectors in Norman Alber´s notation.

 

For more information see: http://en.wikipedia.org/wiki/Del

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I can't say I fully understand, but I do have an appreciation for why the current, magnetic field and force are all at right angles to each other now, and I thank you all, and I am very grateful for your comments:-) .

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In this calculus we are adding together the way the field is changing in the three directions. Then we say, this change is caused by something. This operation is called divergence, as in the above statement of 'E' field. You can take different directions of the field and ask how they change in the other directions, and that is called curl. Divergence is adding up three components of a field change along their respective directions.

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