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Q.
What is the $SI$ unit of permeability?
Physical World, Units and Measurements
Solution:
$L_{\text {solenoid }}=\frac{\mu_{0} N^{2} A}{l},$ where $N$ is the number of turns
$\Rightarrow \left[\mu_{0}\right]=\frac{[L][l]}{\left[N^{2}\right][A]}=\frac{\text { henry } \cdot m }{ m ^{2}}$
$\left[\mu_{0}\right]=$ henry per metre.
Also, $ B_{\text {solenoid }}=\frac{\mu_{0} N I}{l}$
$\left[\mu_{0}\right]=\frac{[B][l]}{[N][I]}=$ tesla metre per ampere.
Also, $\phi=B A \Rightarrow $ weber $=$ tesla metre $^{2}$
$Wb = Tm ^{2}$
As $\left[\mu_{0}\right]= TmA ^{-1}= Wb \,m ^{-1} A ^{-1}=$ Weber per (ampere
metre).