$$\nabla \times A = B$$
$A$ is vector magnetic potential, $\mathrm{Wb/m}$
$B$ is magnetic field intensity, $\mathrm{Wb/m^2}$
Where does one more m come from for $B$? Is that from the gradient operator so it is in meter or something?
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2 Answers
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The del operator is
$$\nabla = [\frac{d}{dx},\frac{d}{dy},\frac{d}{dz}] $$
and ends up taking derivatives of A with respect to x,y,z axis, which are displacements measured in meters.
Since a derivative is a rate of change, you are finding the change in A(Wb/m) per meter, hence $$\frac{Wb/m}{m}=Wb/m^2$$.
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1$\begingroup$ note that the gradient operator refers to a scalar potential. here, you're dealing with a vector field and are taking its curl $\endgroup$ Commented Apr 5, 2021 at 1:40
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$\begingroup$ @user256872 Oh yes, I tend to call it gradient since that's the only operation between gradient, curl and divergence where it appears by its lonesome. What is the name for just that operator again? Del? $\endgroup$– DKNguyenCommented Apr 5, 2021 at 1:45
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$\begingroup$ at first I did not include the name because I forgot it too :P but after looking it up it's indeed the Del operator $\endgroup$ Commented Apr 5, 2021 at 1:57
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As you suggested the del operator provides the additional “per/meter”. The del is the rate of change with respect to position. So just like taking a derivative with respect to time divides the units by time so also taking a derivative with respect to position divides the units by distance.
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2$\begingroup$ Looking at the electrical analogue is also helpful... $-\vec \nabla V=\vec E$, where $V$ has units of $\rm Volts$ and $\vec E$ has units of $\rm Volts/m$. $\endgroup$– robphyCommented Apr 5, 2021 at 2:46