4.27. (7 points) a sheet lying on the z = -2 cm to z = 2 cm. how much torque (magnitude) does it experience…

Explanation:

Step1: Recall Biot - Savart law for magnetic field intensity

For a current - density $\vec{K}$, the infinitesimal magnetic field intensity $d\vec{H}$ is given by $d\vec{H}=\frac{1}{4\pi}\frac{\vec{K}\times d\vec{l}}{r^{2}}$. Here, we consider a surface - current problem. In the given problem, for a point on the z - axis $(0,0,z)$, and for a surface - current density $\vec{K}$ in the xy - plane, we use the fact that $d\vec{l}=dxdy\hat{n}$ (where $\hat{n}$ is the unit normal to the surface element).

Step2: Calculate the cross - product for $d\vec{H}$

Let $\vec{K}=\frac{I}{2\pi}\frac{x\hat{a}{x}+y\hat{a}{y}}{x^{2}+y^{2}}$ for $x\geq0,y\geq0$ and $0$ otherwise. For a point $(0,0,z)$ on the z - axis, the position vector from a point $(x,y,0)$ in the xy - plane to the point on the z - axis is $\vec{r}=x\hat{a}{x}+y\hat{a}{y}-z\hat{a}{z}$. Then $\vec{K}\times\vec{r}=\frac{I}{2\pi}\frac{\left|\begin{array}{ccc}\hat{a}{x}&\hat{a}{y}&\hat{a}{z}\x&y&0\x&y& - z\end{array}\right|}{x^{2}+y^{2}}=\frac{I}{2\pi}\frac{-z y\hat{a}{x}+z x\hat{a}{y}}{x^{2}+y^{2}}$. And $d\vec{H}=\frac{1}{4\pi}\frac{\vec{K}\times\vec{r}}{r^{2}}dxdy$, where $r = \sqrt{x^{2}+y^{2}+z^{2}}$. Converting to polar coordinates $x = \rho\cos\varphi$, $y=\rho\sin\varphi$, $dxdy=\rho d\rho d\varphi$, and $\vec{K}=\frac{I}{2\pi}\frac{\rho\cos\varphi\hat{a}{x}+\rho\sin\varphi\hat{a}{y}}{\rho^{2}}=\frac{I}{2\pi\rho}(\cos\varphi\hat{a}{x}+\sin\varphi\hat{a}{y})$ for $\rho\geq0,0\leq\varphi\leq\frac{\pi}{2}$. The position vector from $(\rho\cos\varphi,\rho\sin\varphi,0)$ to $(0,0,z)$ is $\vec{r}=\rho\cos\varphi\hat{a}{x}+\rho\sin\varphi\hat{a}{y}-z\hat{a}{z}$. Then $\vec{K}\times\vec{r}=\frac{I}{2\pi}\frac{\left|\begin{array}{ccc}\hat{a}{x}&\hat{a}{y}&\hat{a}{z}\\rho\cos\varphi&\rho\sin\varphi&0\\rho\cos\varphi&\rho\sin\varphi& - z\end{array}\right|}{\rho}=\frac{I}{2\pi}\frac{-z\rho\sin\varphi\hat{a}{x}+z\rho\cos\varphi\hat{a}{y}}{\rho}=\frac{I z}{2\pi}(-\sin\varphi\hat{a}{x}+\cos\varphi\hat{a}{y})$. And $d\vec{H}=\frac{1}{4\pi}\frac{\vec{K}\times\vec{r}}{(\rho^{2}+z^{2})}\rho d\rho d\varphi$.

Step3: Integrate to find $\vec{H}$

(a) [ \begin{align*} d\vec{H}&=\frac{I z}{8\pi^{2}}\frac{-\sin\varphi\hat{a}{x}+\cos\varphi\hat{a}{y}}{(\rho^{2}+z^{2})}\rho d\rho d\varphi\ \vec{H}&=\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}\frac{I z}{8\pi^{2}}\frac{-\sin\varphi\hat{a}{x}+\cos\varphi\hat{a}{y}}{(\rho^{2}+z^{2})}\rho d\rho d\varphi \end{align*} ] First, integrate with respect to $\rho$: $\int_{0}^{\infty}\frac{\rho}{(\rho^{2}+z^{2})}d\rho=\frac{1}{2}\ln(\rho^{2}+z^{2})\big|{0}^{\infty}$. We can also use the substitution $u = \rho^{2}+z^{2}$, $du = 2\rho d\rho$. Then $\int{0}^{\infty}\frac{\rho}{(\rho^{2}+z^{2})}d\rho=\left[\frac{1}{2}\ln(u)\right]{z^{2}}^{\infty}$. Another way is to use the formula $\int\frac{\rho}{(\rho^{2}+z^{2})}d\rho=\frac{1}{2}\frac{1}{z}\tan^{- 1}(\frac{\rho}{z})\big|{0}^{\infty}=\frac{\pi}{4z}$. Integrating with respect to $\varphi$: $\int_{0}^{\frac{\pi}{2}}(-\sin\varphi)d\varphi = 1$ and $\int_{0}^{\frac{\pi}{2}}\cos\varphi d\varphi = 1$. [ \begin{align*} \vec{H}&=\frac{I}{8\pi z}(\hat{a}{x}+\hat{a}{y}) \end{align*} ]

(b) The expression for $\vec{H}$ at any point on the z - axis is $\vec{H}=\frac{I}{8\pi z}(\hat{a}{x}+\hat{a}{y})$ for $z\neq0$.

Answer:

(a) $d\vec{H}=\frac{I z}{8\pi^{2}}\frac{-\sin\varphi\hat{a}{x}+\cos\varphi\hat{a}{y}}{(\rho^{2}+z^{2})}\rho d\rho d\varphi$ (in polar coordinates for $x\geq0,y\geq0$) (b) $\vec{H}=\frac{I}{8\pi z}(\hat{a}{x}+\hat{a}{y})$ for $z\neq0$