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, $e=-\frac{\Delta |}{\Delta t}$ where $\Delta |$ 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 $E=\frac{\text{dimensions}\text{of}\text{W}}{\text{dimensions}\text{of}\text{q}}=\frac{[M{L}^2{T}^{-2}]}{[Q]}$ $=[M{L}^2{T}^{-2}{Q}^{-1}]$