I am curious about estimating power losses due to eddy currents. Looking on Wikipedia I find an expression for power dissipation under limited circumstances, $$ P = \frac{\pi^2 B^2 d^2 f^2}{6k\rho D} $$ where $P$ is the power in watts per kilogram, $B$ is the peak field, $d$ is the thickness of the conductor, $f$ is the frequency, $k\sim1$ is a dimensionless constant which depends on the geometry, $\rho$ is the resistivity, and $D$ is the mass density.
However, I can't make the units work out. The tricky ones are usually the electromagnetic units. From the Lorentz force $\vec F = q\vec v\times\vec B$ I find $$ \mathrm{ 1\,T = 1 \frac{N\cdot s}{C\cdot m} }. $$ From Ohm's law $V=IR$, $$ \mathrm{ 1\,\Omega = 1\,\frac{V}{A} = 1\,\frac{N\cdot s}{C^2} }, $$ and the dimension of $\rho$ is $\mathrm{\Omega\cdot m}$.
So the dimensions of ratio $P$ should be \begin{align*} \left[ B^2 (d\,f)^2 \rho^{-1} D^{-1} \right] &= \mathrm{ \left( \frac{N\cdot s}{C\cdot m} \right)^2 \left( \frac ms \right)^2 \left( \frac{C^2}{N\cdot s\cdot m} \right) \left( \frac{m^3}{kg} \right) }\\\ &= \mathrm{ \left( \frac{N^2}{C^2} \right) \left( \frac{C^2}{N\cdot s\cdot m} \right) \left( \frac{m^3}{kg} \right) }\\ &= \mathrm{ \left( \frac{N}{m\cdot s} \right) \left( \frac{m^3}{kg} \right) } = \mathrm{ \frac{N\cdot m^2}{s\cdot kg} = \frac{W\cdot m}{kg} } \end{align*} This is different from the stated units of $\mathrm{W/kg}$. Am I making some stupid mistake? Is the formula wrong? What's happening here?