Question

By using mass[M] , length[L], time[T] and current[A], find the dimension of permeability?

Answer 1

Task. By using mass[M], length[L], time[T], and current[A], find the dimension of permeability.

Solution.

Answer. Permeability is the measure of the ability of a material to support the formation of a magnetic field within itself. It is denoted by letter $\mu$ and measured in $[N]\cdot[A]^{-2}$, where [N] means “newton” and [A] means “ampere”. We should express $[N]\cdot[A]^{-2}$ via [M], [L], [T] and [A].

Notice that [N] the measure for the force $F$. Using newton law $F=ma$, where $m$ is the mass and $a$ is the acceleration measured in $m/s^{2}$ we see that

$[N]=[M]\cdot[L]/[T]^{2}=[M]\cdot[L]\cdot[T]^{-2}$

Hence the measure for permeability can be written as follows:

$[N]\cdot[A]^{-2}=[M]\cdot[L]\cdot[T]^{-2}\cdot[A]^{-2}.$

Answer. $[M]\cdot[L]\cdot[T]^{-2}\cdot[A]^{-2}$.