Question

The dimensional formula for emf e in MKS system will be

• A. $ \left[ M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}} \right] $
• B. $ \left[ M{{L}^{2}}{{T}^{-1}} \right] $
• C. $ \left[ M{{L}^{-2}}{{Q}^{-1}} \right] $
• D. $ [ML{{T}^{-2}}\,{{Q}^{-2}}] $

Answer 1

Answer: (A)

From Faradays law of electromagnetic induction, $ e=-\frac{\Delta \text{ }\!\!|\!\!\text{ }}{\Delta t} $ where $ \Delta \text{ }\!\!|\!\!\text{ } $ is magnetic flux and At is time. $ \therefore $ Dimensions of $ e=-\frac{\text{dimenstions}\,\text{of}\,\text{magnetic}\,\text{flux}}{\text{dimensions}\,\text{of}\,\text{time}} $ $ =-\frac{[M{{L}^{2}}{{T}^{-2}}{{A}^{-1}}]}{[T]}=[M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}}] $ Where $ \theta $ is charge. Alternative: If during the flow of q coulomb of charge in an electric circuit, the energy supplied by cell is W joule, then the emf E of the cell is $ E=\frac{W}{q}J/C $ Dimensions of $ E=\frac{\text{dimensions}\,\text{of}\,\text{W}}{\text{dimensions}\,\text{of}\,\text{q}}=\frac{[M{{L}^{2}}{{T}^{-2}}]}{[Q]} $ $ =[M{{L}^{2}}{{T}^{-2}}{{Q}^{-1}}] $