Practice portal › Electromagnetic Induction › Self-Inductance
Asked in GSEB Board July 2018 · Units and dimensions
Given: the self inductance L of a coil.
Idea: Φ=LI, so L=Φ/I -- the flux linked per unit current.
[Φ]=[ε][t]=M¹L²T⁻³A⁻¹×T¹=M¹L²T⁻²A⁻¹.
Dividing by [I]=A¹ gives [L]=M¹L²T⁻²A⁻².
A second check: L/R has to be a time, and M¹L²T⁻²A⁻² divided by [R]=M¹L²T⁻³A⁻² is indeed T¹.
So self inductance has the dimensions M¹L²T⁻²A⁻².
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