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Asked in GUJCET 2014 · Units and dimensions of flux
Given: magnetic flux Φ_B.
Idea: Φ_B=BA, and by Faraday's law ε=-(dΦ_B)/(dt), so the flux is (emf) × (time).
[ε]=[V]=M¹L²T⁻³A⁻¹.
So [Φ_B]=M¹L²T⁻³A⁻¹×T¹=M¹L²T⁻²A⁻¹.
Check with Φ_B=BA: [B]=M¹L⁰T⁻²A⁻¹ and [A]=L² give the same thing.
So magnetic flux has the dimensions M¹L²T⁻²A⁻¹ -- in SI the weber, V s.
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