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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?

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Very good practice to double check Wikipedia, but in this case Wikipedia is right. Those E&M units are tricky. From

$F=Eq$ we get that $V=\frac{N \cdot m}{C}$ So $\Omega = \frac{N \cdot s \cdot m}{C^2}$ which solves your problem.

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  • $\begingroup$ I usually remember that $\mathrm{V = J/C}$ relates potential and potential energy. I got my wires crossed and wrote it the same wrong way until it looked right. Thanks! $\endgroup$
    – rob
    Commented May 31, 2014 at 0:33
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There is an error in the above: "From Ohm's law V=IR", Ω=V/A=(J/C)/(C/sec)=(Nm/C)/(C/sec)=Nmsec/CC, not Nsec/C*C

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  • $\begingroup$ Welcome to Physics! This seems to be a duplicate of the accepted answer, except the accepted answer argues that the question used the wrong definition of the volt, while this answer just asserts so. $\endgroup$
    – rob
    Commented Nov 13, 2019 at 15:05

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