Is a Magnetic Field B with Curl B = 0 Possible?

[naftali]

Hi,

Is it possible to have a magnetic field B which has curl B = 0 in all space?
intuitivly such a field will be in a constant direction (like the electric field of an infinite charged plate ) and magnetic field "don't behave" like that, they make circles around currents, but this is not rigorous enough..
I saw a solution in which after getting that curl B = 0 in all space, it was claimed that B = 0 in all space. Is that claim true?

Thanks

 

[Ben Niehoff]

In the absence of electric fields,

\nabla \times \vec B = \mu_0 \vec J

So in fact, at every point in space where there is no current, the curl of B is zero! If you have a line current, the B field will loop around the wire in circles, but the curl is actually zero everywhere, except on the wire itself (where it spikes to infinity, because the current density is also infinite).

If you have a sheet current on an infinite plane, you will get B field lines that are absolutely straight, parallel to the plane, and their density is independent of distance from the plane. In this case, again, the curl is zero everywhere except on the plane itself, where it is infinite.

 

[arunma]

Sure it's possible. Consider Ampere's Law with the displacement current term:

\vec{\nabla} \times \vec {B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \dfrac{d\vec{E}}{dt}

If the current density is zero and the electric field is static, then you can absolutely have a curl-free magnetic field. Unfortunately, you typically get the trivial solution of a zero magnetic field (except in the cases that Ben specified, where the curl is a Dirac delta function). Mathematically speaking, magnetic field lines have no sources or sinks and always loop back on themselves, so you can simply say that the field lines begin and end at infinity. In reality though, I can't think of any physical case where this would happen. That said, you can certainly have magetic fields that are locally uniform and which have zero curl. One way to produce such a field is to use Helmholtz coils, a set of two current-carrying wires which have a radius equal to their separation distance. Devices such as NMR also produce very uniform fields with effectively no curl in the region of uniformity.

Obviously you'd never have a physical situation where a magnetic field is produced which is uniform throughout all of space. When physicists say "infinity," we're usually referring to a length much longer than the length scale of the problem we're working with.

 

[manofphysics]

Consider a closed circular loop of radius r about a straight wire carrying a current I ,
What is the line integral around this closed loop? We have chosen a path with constant radius, so the magnetic field at every point on the path is the same: B = \frac{2I}{Rc} . In addition, the total length of the path is simply the circumference of the circle: l = 2Πr. Thus, because the field is constant on the path, the line integral is simply:

lineintegral

B·ds = Bl = \frac{2I}{Rc}(2Πr) = \frac{4*\Pi*I}{c}

This equation, called Ampere's Law, is quite convenient. We have generated an equation for the line integral of the magnetic field, independent of the position relative to the source. In fact, this equation is valid for any closed loop around the wire, not just a circular one

The Curl of a Magnetic Field:

From this equation, we can generate an expression for the curl of a magnetic field. Stokes' Theorem states that:

\ointB·ds = \ointcurl B·da

We have already established that \ointB·ds = \frac{4*\Pi}{c}. Thus:

\ointcurl B·da =\frac{4*\Pi}{c}

To remove the integral from this equation we include the concept of current density, J . Recall that I = \ointJ·da . Substituting this into our equation, we find that

\ointcurl B·da = \frac{4*\Pi}{c}\oint J·da

Clearly, then:

curl B= \frac{4*\Pi*J}{c}

Thus the curl of a magnetic field at any point is equal to the current density at that point. This is the simplest statement relating the magnetic field and moving charges.

 

[naftali]

Hi all,

thank you for all your answers.
Actually, I know Amprer's law.
In the question I talk about the J and e0*dE/dt cancled each other :
a point charge in the origin Q which gives current I=dQ/dt symmetric for all directions.
so we have :
(a) J=I/(4*pi*r^2),
(b) E(r,t) = kQ(t)/r^2, e0*dE/dt = -I/(4*pi*r^2),
So by the corrected Ampere's we have : curl B = mu0(J + e0*dE/dt ) = 0.

Now, is that means B = 0?


P.S
how can I make this nice mathematical symbols you wrote?