Question

The dimensions of self-inductance are:
A) $[ML{T^{ - 2}}{A^{ - 2}}]$
B) $[M{L^2}{T^{ - 1}}{A^{ - 2}}]$
C) $[M{L^2}{T^{ - 2}}{A^{ - 2}}]$
D) $[M{L^2}{T^{ - 1}}{A^{ - 1}}]$

Answer 1

We must know the definition and formula of self-inductance. Substitute the formula with the SI units of all the values and then convert the SI units to fundamental units to get the dimensional formula.

Complete step by step solution:
When the current in a circuit changes, a magnetic field is generated around the circuit. This magnetic field due to a current carrying circuit induces emf in the circuit. This induced emf in the circuit due to the change in current in the same circuit is known as self-induction. This emf is known as self-induced emf.
Mathematically, self-induction is equal to the total magnetic flux divided by the current in the circuit.
This equation is given by
$L = \dfrac{\phi }{i}$
where $L$ is self-inductance
$\phi$ is magnetic flux
$i$ is the current in the circuit
Using this formula now we find the dimensional formula for self-inductance.
Before finding the dimensional formula for self-inductance, understand the meaning of dimensional formula.

The dimensional formula gives the relation between the SI units’ also known as derived units and fundamental units.
The fundamental units mentioned here in the options are given as mass $[M]$ , length $[L]$, time $[T]$ and electric current $[A]$ .
So now we first substitute the formula of self-inductance with the SI units and then convert the SI units to fundamental units.
$L = \dfrac{\phi }{i}$
Substituting magnetic flux $\phi = BA$ where $B$ is magnetic field and $A$ is the area
$\Rightarrow L = \dfrac{{BA}}{i}$
The SI units of $B$ is $Tesla$ , $A$ is ${m^2}$ and $i$ is $Ampere(A)$
${m^2}$ and $Ampere$ are fundamental units so we don’t need to convert them.
We only need to convert $Tesla$ to fundamental units.
$Tesla$ is the SI unit of magnetic field where magnetic field is given by the equation
$F = Bvq$ where $F$ is force (SI unit $Newton$ )
$q$ is charge (SI unit $Coulomb$ )
$v$ is velocity (SI unit $m{s^{ - 1}}$ )
$\therefore B = \dfrac{F}{{vq}}$
$\Rightarrow B = \dfrac{N}{{Cm{s^{ - 1}}}}$
Here, substituting $C{s^{ - 1}} = Ampere$ as $q = \dfrac{I}{t}$
$\Rightarrow B = \dfrac{N}{{Am}}$
Substituting $N = kgm{s^{ - 2}}$ as $F = ma$
$\Rightarrow B = \dfrac{{kg{s^{ - 2}}}}{A}$
Substituting these units of $B$ in $L = \dfrac{{BA}}{i}$
$\therefore L = \dfrac{{kg{m^2}{s^{ - 2}}}}{{{A^2}}}$
$\Rightarrow L = [{M^1}{L^2}{T^{ - 2}}{A^{ - 2}}]$
Where $M$ defines mass $L$ defines length $T$ defines time and $A$ defines electric current.

$\therefore$ Option $(C), [M{L^2}{T^{ - 2}}{A^{ - 2}}]$ is the right option for the dimensional formula of self-inductance.

Note: We must know which units are fundamental units and which are SI units. Knowing the formulae and relations between the units is important to get the right answer.
Inductance is of two types: self-inductance and mutual-inductance. In self-inductance the change in current and induced emf is in the same circuit whereas in mutual inductance the change in current is in one circuit and the emf is induced in the neighbouring circuit.