Does a bend in a DC current carrying conductor radiate EM waves? An accelerated electric charge creates a transverse em wave radiated at the speed of light. But a dc current negotiating any bend in the conductor must undergo accelerating forces to change spatial direction. This must surely generate some, albeit extremely weak, em radiation?
 A: No, a radiated field requires a changing current density, $\vec J$, and a DC current implies $\frac{\partial}{\partial t}\vec J=0$.
If you think about a single charges accelerating around the bend then yes, there are some radiative terms to the Lienard Wiechert fields. But you do not have a single charge, you have a DC current. The various charges cancel out the radiative terms leaving only the non-radiative terms.
This can be seen more easily in Jefimenko's equations:$$ \mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV'$$$$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2} \times \frac{1}{c} \frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV'$$ for DC sources this simplifies to $$ \mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r)  \right] dV'$$$$\mathbf{B}(\mathbf{r}, t) = -\frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3} \times \mathbf{J}(\mathbf{r}', t_r) \right] dV'$$
Notice that for the DC Jefimenko's equations both the E field and the B field fall off as $|\mathbf{r}-\mathbf{r'}|^{-2}$ which means that the energy stays strongly localized near the sources. It does not radiate.
A: The power radiated by a charge with constant acceleration is
$$P=\frac{q^2}{4\pi\epsilon_0}\frac{a^2}{c^3}.$$ For an electron going around a circular bend of radius (let's say) one centimeter the acceleration is
$$a=\frac{v_d^2}{r}$$ where $v_d$ is the drift velocity for an electron that's around a millimeter per second. Once you put all the numbers together the power radiated by a single electron in this situation comes out to around $10^{-61}$ watts while the kinetic energy of the electron is around $10^{-37}$ J, so we're talking about more than $10^{16}$ years for the electron to radiate away all its energy. This is all to say that the radiation emmited is negligible.
