# Applying Ampere's Circuital Law to a bar magnet

The magnetic field along the axis of a long solenoid comes out to be $$\mu_0nI$$ using ampere's circuital law, considering a rectangular amperian loop.

But if I try to apply the law to an equivalent bar magnet using the same loop, there is no electric current through the loop, but $$\int \overrightarrow B. \overrightarrow {dl}$$ is non zero for $$\overrightarrow B$$ along the magnet's axis. What am I missing? Is the law even applicable here?

The Ampere's law applies to magnetic H field, not magnetic B field because $$\nabla\times\vec{H}=\vec{J}$$, where $$\vec{J}$$ is conducting current. The $$\mu_0nI$$ in the first paragraph of your statement should be changed to $$nI$$. If there is only a magnet and the conduction current is 0, then $$\nabla\times\vec{H}=0$$, so $$\oint\vec{H}\cdot d\vec{l}=0$$, which is consistent with the Ampere's law.
For a bar magnet, there is a 'bound surface current' on its surface given by $$\bf K_{\rm bound}=M\times{\hat n}$$. This acts like a current in a solenoid.
• I am aware of your written $K_\text{bound}$. I have already provided an example calculation here. Maxwell's equations including Ampere's law can be applied without any problem when there are magnets Jul 27, 2022 at 22:03