0

I'm struggling to understand why the curl of this magnetic vector field would be zero:

$$B(x,y,z)=\left(\frac{-y}{(x^2+y^2)},\frac{x}{(x^2+y^2)},0\right)$$

This cartesian field was derived from the equation for a long, straight current carrying wire, with current flowing on the Z axis. I found this derivation here. Note that I removed the constants for simplicity ($\frac{\mu_0*I}{2\pi}$)

If I plot this field (shown in x-y Plane), it looks obvious that it should have curl; since the magnitude of the field lines increase approaching the center, any point in this graph should "spin" clockwise. I realize that this is an intuitive definition of curl, so please correct me if I'm wrong for using it. enter image description here

Also, since this equation was derived from a magnetic field equation, barring any computation error, this HAS TO have curl from Ampere's/Maxwell's Law: $$\nabla X B=\frac{1}{c}\left(4\pi J + \frac{dE}{dt}\right)$$ In the constant current example, $\frac{dE}{dt}=0$, but current density $J$ is proportional to current and distance from the wire.

However, when I calculate the curl of this it results with zero:

$$[-2*\frac{x^2}{(x^2 + y^2)^2} - 2*\frac{y^2}{(x^2 + y^2)^2} + \frac{2}{(x^2 + y^2)}]\hat k=0$$

If anyone could help out mistakes or poor assumptions made here, I would greatly appreciate it!

jng224
  • 3,697
Alex
  • 1
  • The current density is not proportional to the distance from the wire - the current density which yields that magnetic field is zero everywhere except the origin, where it is a delta function. – J. Murray May 17 '21 at 19:03
  • @J. Murray: I see your point about the current density -- I believe I made the mistake of using average current. To clarify though, the B field (flux density) wouldn't be zero outside the wire? You can directly measure the field/flux about a current flow. – Alex May 17 '21 at 19:17
  • Does this answer your question? Why is this vector field curl-free? See, in particular, Diracology's answer (currently the second on the page.) – Michael Seifert May 17 '21 at 19:19
  • 1
    Also, welcome to [Physics.SE]! If this question does get closed as a duplicate, please don't take it personally; it has happened to all of us. This is a well-thought-out question, and I encourage you to stick around and ask more such questions. – Michael Seifert May 17 '21 at 19:22
  • Your calculation of the curl gives $0/0$, not $0$, when $x$ and $y$ are $0$. – G. Smith May 17 '21 at 19:31

0 Answers0