# Ampere's law: what if we have a magnetic field but no current

Gold Member

## Main Question or Discussion Point

If I have a magnetic field describing a closed path it means that this closed path is surrounding a current, right? But if I have no current, is it the displacement current ?
I'm thinking of a magnetic dipole, its magnetic field describes closed paths from its north to south pole, so is there a displacement current surrounding the longitudinal axis of a magnetic dipole?

Related Other Physics Topics News on Phys.org
Dale
Mentor
I'm thinking of a magnetic dipole, its magnetic field describes closed paths from its north to south pole, so is there a displacement current surrounding the longitudinal axis of a magnetic dipole?
No, it isn’t a displacement current. It is what is called a bound current.

Homework Helper
Gold Member
A single loop of current is a magnetic dipole.$\\$ In the case of a solid cylindrical magnet of finite length, in one model used to describe the magnetic effects, the magnetism $\vec{M}$ in the material results in bound magnetic surface currents which are geometrically equivalent to the currents of a solenoid.

Gold Member
Wow, never heard of it.
Is it also given by $\epsilon_0\dfrac{\partial \Phi_E}{\partial t}$ ?

Homework Helper
Gold Member
The magnetic surface currents are presented in Griffiths' E&M textbook. He presents them so quickly, without extra emphasis, that I think many students overlook them. He does a derivation involving the vector potential $A$ for an arbitrary distribution of magnetic dioples, and then at the very end, he shows this is equivalent to having bulk currents with current density $J_m=\nabla \times M$, along with surface current per unit length $K_m=M \times \hat{n}$. $\\$ (i.e. you get the same $A$ as you would from current sources that are the bulk current from the magnetization along with the bound surface currents, because $A(x)=\frac{\mu_o}{4 \pi} \int \frac{J(x')}{|x-x'|} \, d^3x'$ from current sources $J$). $\\$ See also: https://www.physicsforums.com/threads/magnetic-field-of-a-ferromagnetic-cylinder.863066/
and
https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/ $\\$ And no, what you have shown $I_D=\mu_o \epsilon_o \frac{\partial{\Phi_E}}{\partial{t}}$ is the displacement current that can be found in a capacitor that is charging, as well as in transverse electromagnetic waves.

Last edited:
Gold Member