Question

The dimensions of emf in MKS is
(A) $M{L}^{-1}{T}^{-2}{Q}^{-2}$
(B) $M{L}^2{T}^{-2}{Q}^{-2}$
(C) $ML{T}^{-2}{Q}^{-1}$
(D) $M{L}^2{T}^{-2}{Q}^{-1}$

Answer 1

Hint: The dimensional formula of emf i.e., electromotive force can be calculated using the formula $e={\displaystyle \frac{W}{Q}}$. This is the emf of the cell, i.e., the voltage developed by any source of electrical energy such as battery or dynamo.

Complete step by step solution:
To calculate the dimensions of emf in MKS we use following formula
$e={\displaystyle \frac{W}{Q}}$ …..(1)
Where W = Work done
g = Charge

We must write each physical quantity used in its dimensional form.
W = Fd …..(2)
Where F = force, d = displacement
So, F = ma ….. (3)
Where m = mass, a = acceleration
Mass m = $[M]$ …..(4)
Acceleration $a={\displaystyle \frac{ch\arg einvel}{time}}$
vel $V=[L{T}^{-1}]$ …..(5)
Time $t=[T]$ …..(6)
From (4), (5), (6) $\to$ (3)
$F=[M]{\displaystyle \frac{[L{T}^{=1}]}{[T]}}=[ML{T}^{-2}]$ …...(7)
Displacement $d=[L]$ …...(8)
Put values from (7), (8) $\to$ equation (2)
$W=[ML{T}^{-2}][L]=[M{L}^2{T}^{-2}]$ …..(9)
Now charge $Q=[Q]$ ……(10)
Or we can take charge as coulomb also
Put values from equation (9), (10) $\to$ (1)
$e={\displaystyle \frac{[M{L}^2{T}^{-2}]}{Q}}$
$\boxed{{\displaystyle e=[M{L}^2{T}^{-2}{Q}^{-1}]}}$

Hence, option (D) is correct.

Note: Another method for solving this question would be by using the formula for induced emf i.e.,
$e={\displaystyle \frac{d{\varphi }_B}{dt}}$
Where ${\varphi }_B=$ Magnetic flux
${\varphi }_B=BA$
Where B = Magnetic field
A = Area
So, $e={\displaystyle \frac{d(BA)}{dt}}$
If N is no. of turns in coil then $e=N{\displaystyle \frac{d(BA)}{dt}}$

While solving dimensional formula questions students must note that energy physical quantity must be expressed in its absolute units only.