# Question regarding biot savart law and magnetic fields from current

1. Aug 5, 2012

### bigerst

say you have a very very long (or infinite) straight wire the carried a charge (glued to the wire) and the wire moves forward with speed v with respect to reference frame S, this creates a current, which according to the Biot Savart law creates a magnetic field. But in reference frame S' relatively stationary to the wire since the charges are not moving there is no current and hence there is no magnetic field. suppose a statinary charge is placed at the origin of S', in S' the charge experiences no magnetic force but in S it does?
i havent really done special relativity yet so can someone resolve this in terms of classical mechanics (assuming v <<c)?

2. Aug 5, 2012

### Philip Wood

Is your worry that the same charge can experience different forces measured in different reference frames? In non-relativistic mechanics (Galilean transform) force is the same in inertial frames moving at constant relative velocity. So you can't, imo, resolve your problem without using Special Relativity theory – when it all comes clear.

May have misunderstood your problem. Others are bound to respond: this topic is a perennial favourite.

3. Aug 5, 2012

### cpsinkule

if you have 0 relative motion with respect to the charges in the wire you will experience no magnetic field. if the relative motion is not 0, you will experience a magnetic field

4. Aug 5, 2012

### dydxforsn

You're correct bigerst, the moving frame (S') will not have a magnetic field present. The fact that the very existence of the magnetic field depends on relative velocities screams out special relativity. The big result from applying special relativity to electromagnetism is the uniting of the electric and magnetic fields as one in the same object that transforms from one moving frame to another in a specific way (the "object" that contains both the electric and magnetic field that "unites" the two is called the electromagnetic field tensor, but I don't really want to get into it because of the mathematical sophistication. Just know that it has six components and obeys the 6 transformation equations I list at the end of this post.) In fact, if you allow the S' frame to be moving at constant velocity in the 'x' direction, you get the following equation that relates the fields in the S (stationary) frame to the S' (frame traveling in uniform translational motion in the 'x' direction):

$\begin{split} {E}_{x'} &= E_{x} \\ {E}_{y'} &= \gamma (E_{y} - v B_{z}) \\ {E}_{z'} &= \gamma (E_{z} + v B_{y}) \\ {B}_{x'} &= B_{x} \\ {B}_{y'} &= \gamma (B_{y} + \frac{v}{c^2} E_{z}) \\ {B}_{z'} &= \gamma (B_{z} - \frac{v}{c^2} E_{y}) \end{split}$

Here 'v' is the velocity and $\large \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$.

You can sort of "play" around with these transformation equations to get a better feel of what they really mean. Specifically, look what happens when you have no magnetic field in the stationary frame (the un-primed frame) and you want to know what kind of magnetic field you gain in a moving frame (the primed frame). You can only gain a magnetic field that is transverse to the direction of motion (because $B_{x'} = B_{x}$ and $B_{x} = 0$). Just things like that are fun to notice (you can thus see why the Biot-Savart law produces field in the directions it does..)

Edit*** I guess I should say that to respond to your apparent "discrepancy", if you place a charge in the S' frame so that it's stationary according to an observer in the S' frame, it will indeed produce a magnetic field for the stationary S frame. The magnetic force will be different, but so will the force from the electric field!

Also this can be explained "classically" from Maxwell's equations I'm sure (though special relativity is implicit in them anyway..), but I haven't thought of how to do this yet. I may come back to this.

Edit again*** Yeah, you can just say that there are transverse planes traveling along with the moving charge that experience a changing flux because the charge is moving closer or further from those planes at speed 'v'. Changing fluxes produce fields from Faraday and Ampere's Laws:

$\begin{split} \nabla \times \vec{E} &= - \frac{{\partial}{\vec{B}}}{{\partial}{t}} \\ \nabla \times \vec{B} &= \mu_{o} \vec{J} - \mu_{o} \epsilon_{o} \frac{{\partial}{\vec{E}}}{{\partial}{t}} \end{split}$

That's a "classical" explanation..

Last edited: Aug 5, 2012
5. Aug 6, 2012

### bigerst

thanks. it helps

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