Time-varying current in a long straight wire

[AxiomOfChoice]

I know the B field due to a current-carrying wire is

$$\frac{\mu_0 I}{2\pi r}$$

But suppose the current in the wire is not constant, but varies in time. Maybe sinusoidally: $I(t) = I\cos \omega t$. Do we have

$$B(t) = \frac{\mu_0 I\cos \omega t}{2\pi r}?$$

 

[AxiomOfChoice]

(Of course, I think the answer to the above question is YES, but I can't convince myself I'm correct.)

 

[Bob S]

In coaxial cables and other transmission lines, the signal voltage drop per unit length is given by the inductance L per unit length, and the current by the shunt capacitance C per unit length. The net result is given by the so-called telegraph equations. See
http://www.ipm.virginia.edu/process/PVD/Pubs/thesis4/chapter3.pdf
The signal has a characteristic impedance Z = sqrt(L/C) and a velocity v= 1/sqrt(LC), which is typically 0.66 to 0.9 times the speed of light. In a coaxial cable, the B field is azimuthal (as you point out), and is confined to the space between the conductors, as is the radial electric field.
[added] Note that although both L and C are reactive impedances, their product and ratio are both real. So both Z = sqrt(L/C) and velocity = 1/sqrt(LC) are real.
Bob S.

 

[AxiomOfChoice]

Thanks for your response, Bob. But is the answer to my question "Yes" or "No"?

I'm just talking about an ideal thin wire in a vacuum that carries a sinusoidal current.

 

[Bob S]

Thanks for your response, Bob. But is the answer to my question "Yes" or "No"?

I'm just talking about an ideal thin wire in a vacuum that carries a sinusoidal current.

Where is your return current? The voltage on the wire has radial field lines. Where do they terminate?

B(t) = (u₀I/2 pi r) cos(wt - kx)

The wave shape travels down the wire at about 20 to 30 cm per nanosecond.

 

[vanesch]

The answer to the question of the OP is "yes, but". It is a good approximation as long as the sinusoidal variation is "slow". What does that mean ? Consider the period of the sinusoidal variation, and consider the wavelength of an EM wave that goes with it. Let us call it lambda.

As long as the diameter of the wire is WAY smaller than lambda, and the place where you want to consider the B-field is at a distance from the wire much smaller than lambda, your formula is still a good approximation.

People call this region also the "near field" (although "near field" might mean a larger part than what I just described).

For instance, if you have 50 Hz (or 60 Hz) current flowing through a wire of say, 1 km long, your formula works very well for all practical purposes.

If you have a 1 GHz current flowing through a cable of 2 meters long, and you're looking for the field at 20 cm, you're way off with that formula.