Question 7.8: If plane z = 0 carries uniform current K = Ky ay, H = { (1/2...

Consider the current sheet as in Figure 7.21. From eq. (7.42)

A=\int_{S}\frac{\mu_{o}KdS}{4\pi R}      for surface current

dA=\int_{S}\frac{\mu_{o}KdS}{4\pi R}

In this problem, K=K_{y}a_{y},  dS=dx^{'}dy^{'}, and for z\gt 0

R=|R|=|(0,0,z)-(x^{'},y^{'},0)|=[(x^{'})^{2}+(y^{'})^{2}+z^{2}]^{1/2}                                   (7.8.1)

where the primed coordinates are for the source point while the unprimed coordinates are for the field point. It is necessary (and customary) to distinguish between the two points to avoid confusion (see Figure 7.19). Hence

dA=\frac{\mu_{o}K_{y}dx^{'}dy^{'}a_{y}}{4\pi[(x^{'})^{2}+(y^{'})^{2}+z^{2}]^{1/2}}

dB=\nabla \times dA=-\frac{\partial}{\partial z}dA_{y}a_{x}=\frac{\mu_{o}K_{y}zdx^{'}dy^{'}a_{x}}{4\pi[(x^{'})^{2}+(y^{'})^{2}+z^{2}]^{3/2}}

B=\frac{\mu_{o}K_{y}za_{x}}{4\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{dx^{'}dy^{'}}{[(x^{'})^{2}+(y^{'})^{2}+z^{2}]^{3/2}}                                                       (7.8.2)

In the integrand, we may change coordinates from Cartesian to cylindrical for convenience so that

B=\frac{\mu_{o}K_{y}za_{x}}{4\pi}\int_{\rho^{'}=0}^{\infty}\int_{\phi^{'}=0}^{2\pi}\frac{\rho^{'}d\phi^{'}d\rho^{'}}{[(\rho^{'})^{2}+z^{2}]^{3/2}}

=\frac{\mu_{o}K_{y}za_{x}}{4\pi}2\pi\int_{0}^{\infty}[(\rho^{'})^{2}+z^{2}]^{-3/2}1/2d[(\rho^{'})^{2}]

=\frac{\mu_{o}K_{y}za_{x}}{2}\frac{-1}{[(\rho^{'})^{2}+z^{2}]^{1/2}}|_{\rho^{'}=0}^{\infty}=\frac{\mu_{o}K_{y}a_{x}}{2}

Hence

H=\frac{B}{\mu_{o}}=\frac{K_{y}}{2}a_{x},   for  z\gt 0

By simply replacing z by -z in eq. (7.8.2) and following the same procedure, we obtain

H=-\frac{K_{y}}{2}a_{x},   for   z\lt 0