Current sheet

An infinite current sheet can be modelled as an infinite number of parallel wires all carrying the same current. Assuming each wire carries current I, and there are N wires per unit length, the magnetic field can be derived using Ampère's law:

$${\displaystyle \oint _{R}\mathbf {B} \cdot \mathbf {dl} =\mu _{0}I_{\text{enc}}}$$  $${\displaystyle \oint _{R}B\cos(\theta )\,dl=\mu _{0}I_{\text{enc}}}$$ 

R is a rectangular loop surrounding the current sheet, perpendicular to the plane and perpendicular to the wires. In the two sides perpendicular to the sheet, ${\displaystyle \mathbf {B} \cdot d\mathbf {s} =0}$  since ${\displaystyle \cos(90^{\circ })=0}$ . In the other two sides, ${\displaystyle \cos(0)=1}$ , so if S is one parallel side of the rectangular loop of dimensions L × W, the integral simplifies to: $${\displaystyle 2\int _{S}Bds=\mu _{0}I_{\text{enc}}}$$  Since B is constant due to the chosen path, it can be pulled out of the integral: $${\displaystyle 2B\int _{S}ds=\mu _{0}I_{\text{enc}}}$$  The integral is evaluated: $${\displaystyle 2BL=\mu _{0}I_{\text{enc}}}$$  Solving for B, plugging in for I_enc (total current enclosed in path R) as I×N×L, and simplifying: $${\displaystyle {\begin{aligned}B&={\frac {\mu _{0}I_{\text{enc}}}{2L}}={\frac {\mu _{0}INL}{2L}}\\[1ex]&={\frac {\mu _{0}IN}{2}}\end{aligned}}}$$  Notably, the magnetic field strength of an infinite current sheet does not depend on the distance from it.

The direction of B can be found via the right-hand rule.

A well-known one-dimensional current sheet equilibrium is the Harris sheet, which is a stationary solution to the Maxwell–Vlasov system.^[6] The magnetic field profile of a Harris sheet along ${\displaystyle y=0}$  is given by $${\displaystyle \mathbf {B} (y)=B_{0}\tanh \left({\frac {y}{\delta }}\right)\mathbf {\hat {x}} ,}$$  where ${\displaystyle B_{0}}$  is the asymptotic magnetic field strength and ${\displaystyle \delta }$  provides the thickness of the current sheet. The current density is given by $${\displaystyle \mathbf {J} (y)=-{\frac {B_{0}}{\mu _{0}\delta }}\operatorname {sech} ^{2}\left({\frac {y}{\delta }}\right)\mathbf {\hat {z}} .}$$  The plasma pressure is given by $${\displaystyle p(y)={\frac {B_{0}^{2}}{2\mu _{0}}}\operatorname {sech} ^{2}\left({\frac {y}{\delta }}\right)+p_{0},}$$  where ${\displaystyle p_{0}}$  is the asymptotic pressure.