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First of all we all should know the meaning of the phrase "the electron is accelerating." It does not mean that $a=\mathrm{d}v/\mathrm{d}t$. The accelerating electron means that the electron is changing its electric field. Obviously no electromagnetic radiations will be produced during orbital motion of the electron because the electron doesn't radiate ...


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You are on the right way with your line of thought. The only part of the puzzle you are missing is, as you have correctly identified, the explanation of $$\oint{\overrightarrow{r_0} \times d\overrightarrow{F}} = \overrightarrow{r_0} \times\oint{ d\overrightarrow{F}}$$ It is actually very simple: $\overrightarrow{r_0}$ is a constant vector (it doesn't depend ...


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You can extract the constant term out of integral directly, however keep the operation unchanged. See: $$\oint \vec{r}_0\times \mathrm{d}\vec{F}=\oint\epsilon_{ijk}r_{0i}\mathrm{d}F_j\hat{e}_k=\epsilon_{ijk}r_{0j}\oint\mathrm{d} F_j \hat{e}_k=\vec{r}_0\times\oint\mathrm{d}F$$ Similarly: $$\int \vec{B_0}\cdot \mathrm{d}\vec{S}= \vec{B_0}\cdot ...


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This is easy to see if you use the Maxwell equations to arrive at the decoupled, inhomogeneous wave equations for the fields, $$\begin{split}\Box \vec{E} &= - \mu_0 \frac{\partial \vec{J}}{\partial t} - \vec{\nabla} \frac{\rho}{\varepsilon_0},\\ \Box \vec{B} &= \mu_0\vec{\nabla}\times \vec{J}, \end{split} $$ with $\Box \equiv \frac{1}{c^2} ...


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The short answer is yes. H-field can have sources and sinks - these are what the "poles" of a bar magnet are. In an LIH medium $B = \mu_0 \mu_r H$. Even though $\nabla \cdot B = 0$ (always), this does not mean that $\nabla \cdot H =0$ because there is an instantaneous change in $\mu_r$ at the boundaries between media. Lines of H-field begin and end on ...


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Logically speaking, you may have a point. This is because, you have made the distinction that the primed co-ordinates indicates the source point and the non-primed coordinates indicates the evaluation point. However, we know that the homogeneity and isotropy of space implies that the laws of physics are invariant w.r.t spacial translations and rotations. ...


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Since both the balls will be at gravitational equipotential surface, they will fall with same velocity. While the distance for both the balls is same (i.e. they have the same height), time will also be same for both.


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The problem is with your current density $\vec{j}$, let's start with it's basic definition as current over area, $\vec{j} = \frac{Q\vec{v}}{4\pi R^2}$, now knowing that the angular velocity can be used to get the linear velocity of the charge on the spherical surface using $\vec{v}=\vec{\omega}\times R\hat{r} = \omega R \hat{z} \times \hat{r} = \omega R ...



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