The magnetic field produced by an infinitely long wire pointing out of the plane at the origin can be expressed as a complex function $$B(R, \phi) = i\frac{\mu_0I_0}{2\pi}\frac{e^{i\phi}}{R},$$ where $I_0$ is the current and $\mu_0$ is the permeability of free space. I want to calculate this integral: $$I = \oint_{|z|=R}B(R, \phi)dz.$$ From Ampere's law we should find that it simply equal to $\mu_0 I_0$. But I don't know how to show this. Here is what I have tried so far $$z=Re^{i\phi},$$ $$\implies dz = iRe^{i\phi} d\phi.$$ Hence, $$I = \int_{0}^{2\pi} i\frac{\mu_0I_0}{2\pi}\frac{e^{i\phi}}{R} i R e^{i\phi}d\phi.$$ But the problem is this integral evaluates to zero. Do you know where I have gone wrong?
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- $\begingroup$ How did you obtain the expression for the magnetic field? If the current $I$ does not vary with time, then the magnetic field has only an azimuthal component with $$B_{\phi}=\frac{\mu_0 I}{2\pi \rho}$$where $\rho$ is distance from the wire to the observation point. So, the field will not be complex valued. $\endgroup$Mark Viola– Mark Viola2024-12-20 21:57:20 +00:00Commented Dec 20, 2024 at 21:57
- $\begingroup$ Since $$\mathbf{\hat{\phi}} = -\sin(\phi)\mathbf{\hat{x}}+\cos(\phi)\mathbf{\hat{y}}.$$ This is analogous to $$i\exp(i\phi)=-\sin(\phi) + i \cos(\phi),$$ where the real axis corresponds to the $x$-axis and the $y$-axis corresponds to the imaginary axis. $\endgroup$Peanutlex– Peanutlex2025-01-26 10:44:37 +00:00Commented Jan 26 at 10:44
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2 Let's consider the dot product between two vectors $\mathbf{F}$ and $\mathbf{G}$ where $$\mathbf{F}=F_r \mathbf{\hat{x}} + F_i\mathbf{\hat{y}},$$ $$\mathbf{G}=G_r \mathbf{\hat{x}} + G_i\mathbf{\hat{y}}.$$ The dot product is given by $$\mathbf{F}\cdot\mathbf{G}=F_rG_r+F_iG_i.$$ Now consider complex functions $F$ and $G$ where $$F=F_r+iF_i,$$ $$G=G_r+iG_i.$$ Notice that $$FG^* = F_rG_r + F_iG_i + i(F_iG_i - F_rG_i).$$ Hence, the dot product analogy for complex numbers is to take the real part of the product of one function with the complex conjugate of another.
Hence, the integral we should be considering is $$I_2 = \oint_{|z|=R}\Re\left[-i\frac{\mu_0I_0}{2\pi}\frac{e^{-i\phi}}{R}dz\right].$$ Hence using the same substitution from the original post we get $$I_2 = \frac{\mu_0I}{2\pi}\int_0^{2\pi}d\phi=\mu_0I_0,$$ as expected.
- $\begingroup$ Where did the negative signs on $i$ come from? $\endgroup$Mark Viola– Mark Viola2024-12-20 21:58:44 +00:00Commented Dec 20, 2024 at 21:58
- $\begingroup$ I took the complex conjugate. Since we are effectively taking the dot product of $B$ and $dz$. $\endgroup$Peanutlex– Peanutlex2025-01-26 10:41:52 +00:00Commented Jan 26 at 10:41