# Maxwell's Equations problem

the final equation

&nabla;XB(x) = &mu;0j(x)

But this means that the curl of the magnetic field at any point is proportional to the current density at that point.

But take the case of a long straight wire carrying current.

The magnetic field surrounding the wire is circular and hence its curl is everywhere constant in value.

But that means that the current density is everywhere constant in value, even at a million miles away from the wire.

what's with that?

Last edited:

yeah, a million miles from the wire the current density is constant, it's zero. Anywhere outside the wire, the current density is zero.

JMD

pmb
Originally posted by ObsessiveMathsFreak
the final equation

&nabla;XB(x) = &mu;0j(x)

But this means that the curl of the magnetic field at any point is proportional to the current density at that point.

But take the case of a long straight wire carrying current.

The magnetic field surrounding the wire is circular and hence its curl is everywhere constant in value.

Outside the wire the curl vanishes since outside the wire j(x) = 0.

Pete

But the curl doesn't vanish. outside the wire the magnetic field is circular, meaning it has a constant curl.

Alexander
Curl is zero where there is no current, pmb is correct. Curl and integral over extended loop are different quantities. Whan you integrate over loop you have to include sources (currents) if the loop includes them.

Hurkyl
Staff Emeritus
Gold Member
Try computing the curl of a circular field somewhere other than the axis.

pmb

But this means that the curl of the magnetic field at any point is proportional to the current density at that point.

I think I see the problem now. The magnetic field is *not* proportional to current density - the *curl* of the magnetic field is. Sorry I din't note that earlier.

Pmb

Man, you guys call the second derivative "curl"?

blegch

Tom Mattson
Staff Emeritus
Gold Member
Originally posted by KillaMarcilla
Man, you guys call the second derivative "curl"?

blegch

No, the curl is the differential operator:

[nab]&times;

which acts on vector fields. It is not the second derivative.

Oh, right, I think I know what you're talking about now

Sorry, I had Math 126 about two years ago, and haven't used most of it since then (except for the geometric series approximations)

h0 h0, I look like quite the f00l now