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Induced electric field due to a moving wire
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[QUOTE="ELB27, post: 4958376, member: 516388"] The position of the wire is ##s'=vt\hat{y}## (If at time ##t=0## it coincided with the z-axis), the position of the arbitrary point is ##s=x\hat{x}+y\hat{y}## (I neglect the z-coordinate as it is irrelevant to the distance between the point and the infinite wire). therefore the distance between them is ##|s-s'| = \sqrt{x^2+(y-vt)^2}##. Then, ##\vec{E} = -\frac{\partial\vec{A}}{\partial t} = \frac{\mu_0I}{2\pi}\frac{\partial\ln s}{\partial t}\hat{z} = \frac{\mu_0I}{2\pi}\frac{\partial\ln s}{\partial s}\frac{\partial s}{\partial t}\hat{z} = \frac{\mu_0I}{2\pi s}\frac{\partial \sqrt{x^2+(y-vt)^2}}{\partial t}\hat{z} = \frac{\mu_0I}{2\pi s}\frac{(-v)(y-vt)}{\sqrt{x^2+(y-vt)^2}}\hat{z} = -\frac{\mu_0Iv}{2\pi s}\sin\phi\hat{z}## where, by trigonometry, ##\sin\phi = \frac{y-vt}{\sqrt{x^2+(y-vt)^2}}##. Thanks you very much, I like this argument - more straightforward. (Although I think that it is identical to mine above - here the equations are identical to the case where the arbitrary point moves with velocity ##\vec{v}=-v\hat{y}##, right? In either case I should have been more rigorous). [/QUOTE]
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Induced electric field due to a moving wire
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