Physics Asked on May 16, 2021
$$nabla times A = B$$
$A$ 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?
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 $$frac{Wb/m}{m}=Wb/m^2$$.
Answered by DKNguyen on May 16, 2021
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.
Answered by Dale on May 16, 2021
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