1
$\begingroup$

Why magnet is alway dipole Even the atom of the magnetic substance is also dipole , how can small atom be dipole . Pls explain me

$\endgroup$
6
  • 1
    $\begingroup$ This is one of the big questions since the math seems to say that a monopole should be possible, we just can't seem to find or make any. $\endgroup$
    – DKNguyen
    Commented Dec 5, 2021 at 20:30
  • $\begingroup$ @DKNguyen Not the math but the physics says that a monopole is impossible. $\endgroup$
    – my2cts
    Commented Dec 5, 2021 at 20:56
  • $\begingroup$ I mean, the math doesn't say it should be possible, since the condition is the Div B is zero, any one could therefore say anything about any vector field "should be possible" as divergence is a property of any vector field $\endgroup$ Commented Dec 5, 2021 at 21:07
  • $\begingroup$ @jensenpaull I think there's math beyond Maxwell. The weird stuff. $\endgroup$
    – DKNguyen
    Commented Dec 5, 2021 at 21:11
  • $\begingroup$ @my2cts I'm pretty confident you could get a Nobel prize if you could convince people that magnetic monopoles are impossible. You may want to publish ASAP if you have evidence of this. $\endgroup$
    – d_b
    Commented Dec 8, 2021 at 6:50

3 Answers 3

2
$\begingroup$

That there are no magnetic monopoles is mostly an arbitrary choice of coordinates (or gauge).

Maxwell's equations have a 'rotational' symmetry between the electric and magnetic fields, as described in this Stack Exchange question.

Through a 'rotation' of $\alpha$, the divergence (e.g. charge density / monopole density) of the electric and magnetic fields may be made to satisfy: $$ \epsilon_0 \nabla \cdot \vec{E}' = \rho_e' = \rho_e \cos \alpha + \mu_0 \epsilon_0 \rho_m \sin \alpha$$ $$ \frac 1 {\mu_0 } \nabla \cdot \vec{B}' = \rho_m' = \rho_m \cos \alpha - \frac{\rho_e}{\mu_0 \epsilon_0} \sin \alpha $$ Clearly, adding $\pi$ to $\alpha$ makes each of the two densities above negative. By the intermediate value theorem, there is some value of $\alpha$ which will make the electric monopole density zero, and the magnetic non-zero: $$ \alpha = \arctan \left(-\frac{\rho_e}{\rho_m} \frac{1}{\mu_0 \epsilon_0}\right) $$ $$ \nabla \cdot \vec{E}' = 0$$ $$ \frac 1 {\mu_0 } \nabla \cdot \vec{B}' = \rho_m' = -2 \frac{\rho_e}{\mu_0 \epsilon_0} \sin\left( \alpha \right) = \left[1 + \left(\frac{\rho_m}{\rho_e} \mu_0 \epsilon_0 \right)^{2} \right]^{-1/2}$$ Since in the traditional formulation of Maxwell's equations the magnetic charge density is set to zero, $\rho_m = 0$, we may rotate all of the electric monopole to the magnetic by setting $\alpha = \pm \frac{\pi}{2}$. Then: $$ \nabla \cdot \vec{E}' = 0$$ $$ \nabla \cdot \vec{B}' = \pm \frac{2 \rho_e}{\mu_0 \epsilon_0} $$ (You may want to check my math with the exact coefficients)

My point is that asking why there are no magnetic monopoles is asking the wrong question. An arbitrary coordinate transformation can always make it appear that one of the two coupled fields has zero divergence (no monopoles) at any chosen point $\vec{x}$ in space. The right question to be asking is why do all monopoles come with the same (or negative) ratio of electric:magnetic charge?. This is what makes it possible to eliminate one of $\rho_m$ or $\rho_e$ through a change of coordinates.

I feel this is important to address because people can sometimes think that there is something special about the magnetic field having no monopoles. In fact that is a red herring. The fundamental symmetry is a common ratio of monopole charge in all observed particles.

The answer to the good question? I do not know.

$\endgroup$
1
$\begingroup$

In short, the "why" is a tough question, we can only describe reality how we currently see it. Mathematically it is because $\nabla \cdot B = 0$ the magnetic field always creates closed loops as you can only create a curl in the magnetic field. This is an experimental fact

$\endgroup$
0
$\begingroup$

The dipole of permanent magnets is not a mystery if you know that electrons are electric charges and magnetic dipoles in one.

There are two ways to macroscopically enhance these properties:

  • If you separate the electrons, you get a macroscopic electric field around the body of the charge.
  • If you align the magnetic dipoles of the electrons (and the other subatomic particles ), you get a magnet.

The amazing thing about permanent magnets is their ability to maintain the magnetic alignment of the subatomic particles against the action of the thermal motion of the atoms.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.