# Ampere's law, definition of enclosed current

Consider an infinite uniform surface current $$\mathbf{K} = K\mathbf{\hat{x}}$$ flowing over the xy plane. To find the magnetic field $$\mathbf{B}$$ it produces, I first argue that $$\mathbf{B}$$ must solely consist of a y component. To find this component, I consider an Amperian loop that is parallel to the yz plane and that extends an equal distance above and below the xy plane. Applying Ampere's law, I deduce that

$$2Bl = \mu_0I_{\text{enc}}$$

where l denotes the length of the loop in the y direction.

Here comes my question: how do I arrive at $$I_{\text{enc}} = Kl$$? By my logic, if I take $$\mathbf{J}$$ to represent the current flow, then $$\mathbf{J} = 0$$ for $$x\neq 0$$ and $$K\mathbf{\hat{x}}$$ for $$x = 0$$. Thus, since $$I_{\text{enc}}$$ is defined as the flow of $$\mathbf{J}$$ through the surface enclosed by the Amperian loop, and this latter quantity is 0, $$I_{\text{enc}} = 0$$.

I'm feel like I'm messing up something very basic.

• How did you get the LHS in the equation Dec 13, 2021 at 14:34

By my logic, if I take $$\mathbf{J}$$ to represent the current flow, then $$\mathbf{J} = 0$$ for $$x\neq 0$$ and $$K\mathbf{\hat{x}}$$ for $$x = 0$$.
This isn't quite right. The current density $$\mathbf{J}$$ actually is infinite on the surface in order to form a surface current that is non-zero. You should have
$$\mathbf{J} = K\mathbf{\hat{x}}\delta(z)$$