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Advanced Physics Homework Help
Find angular momentum of EM field in terms of q and ##\Phi##
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[QUOTE="WeiShan Ng, post: 6003730, member: 604380"] I try to walk through the steps to see where I have done wrong: Writing the cross products in Cartesian coordinates I have $$[\mathbf{\hat{r}\times (\hat{r}\times \hat{z})}] = \frac{xz}{r^2} \mathbf{\hat{x}} + \frac{yz}{r^2} \mathbf{\hat{y}}+\frac{x^2-y^2}{r^2} \mathbf{\hat{z}}$$ where ##r^2 = x^2+y^2+z^2## So the angular momentum density, ##\mathbf{l}## will be $$\begin{align*}\mathbf{l} &= \frac{QB(x,y)}{4\pi r^3 }(xz \mathbf{\hat{x}}+ yz\mathbf{\hat{y}}+(x^2-y^2)\mathbf{\hat{z}}) \\ &= \frac{QB(x,y)}{4\pi (x^2+y^2+z^2)^{3/2} }(xz \mathbf{\hat{x}}+ yz\mathbf{\hat{y}}+(x^2-y^2)\mathbf{\hat{z}}) \end{align*}$$ x-component and y-component of ##\mathbf{l}## will have the same magnitude but opposite direction at ##\mathbf{l}(x,y,z)## and ##\mathbf{l}(x,y,-z)## for ##\forall (x,y)##. If we integrate it over all volume, they will cancel out and we only left with z-component. The ##\Phi_B## is across xy-plane at z = 0, so it can be written as $$d\Phi = BdA = B r \sin \theta d\phi dr$$ $$\Phi(\theta) = \iint B(r,\theta,\phi) r \sin \theta d\phi dr = \int_{-\infty}^\infty \int^{2\pi}_0 B(r,\theta,\phi) r d\phi dr $$ since ##\theta = \pi/2## at z=0 Integrate ##\mathbf{l}## over all space $$\begin{align*}\mathbf{L} &= \iiint_{all space} \mathbf{l} \, r^2 \sin \theta d\phi dr d\theta \\ &= \int^{\pi}_0 \int^\infty_{-\infty} \int^{2\pi}_0 \frac{-QB \sin^2 \theta}{4\pi r} \mathbf{\hat{z}} \, r^2 \sin \theta d\phi dr d\theta \\ &=\int_0^{\pi} \left[ \int_{-\infty}^{\infty} \int_0^\pi \frac{-QB(r,\theta,\phi)r}{4\pi} d\phi dr \right] \sin^3 \theta d\theta \mathbf{\hat{z}} \\ &= \int_0^{\pi} \frac{-Q\Phi(\theta)}{4\pi} \sin^3 \theta d\theta \mathbf{\hat{z}} \end{align*}$$ And this will be my expression for ##\mathbf{L}##. Hopefully I get it right this time... [/QUOTE]
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Find angular momentum of EM field in terms of q and ##\Phi##
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