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 $e = \dfrac{W}{Q}$. 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
$e = \dfrac{W}{Q}$ …..(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 $a = \dfrac{{ch\arg e\,in\,vel}}{{time}}$
vel $V = [L{T^{ - 1}}]$ …..(5)
Time $t = [T]$ …..(6)
From (4), (5), (6) $ \to $ (3)
$F = [M]\dfrac{{[L{T^{ = 1}}]}}{{[T]}} = [ML{T^{ - 2}}]$ …...(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)
$e = \dfrac{{[M{L^2}{T^{ - 2}}]}}{Q}$
$\boxed{e = [M{L^2}{T^{ - 2}}{Q^{ - 1}}]}$

Hence, option (D) is correct.

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

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