Question

Dimensional formula of self-inductance is:
A. $ML{{T}^{-2}}{{A}^{-2}}$
B. $M{{L}^{2}}{{T}^{-1}}{{A}^{-2}}$
C. $M{{L}^{2}}{{T}^{-2}}{{A}^{-2}}$
D. $M{{L}^{2}}{{T}^{-2}}{{A}^{-1}}$

Answer 1

Hint: Use the expression of magnetic flux to find the dimensional formula of self-inductance. It will be easy if we can reduce the self-inductance equation to the base level. So, we can develop the dimensional formula of magnetic flux from the dimensional formula of force, current and length etc.

Complete step by step answer:
Magnetic flux can be written as,
$\phi =LI$, where $L$ is the self-inductance and $I$ is the current.
So, we can write this equation as, $L=\dfrac{\phi }{I}$
Magnetic flux is the product of magnetic field and area.
$\phi =B\times A$
Magnetic field is the specific area around a magnet at which magnetic force experiences. So,
$B=\dfrac{F}{I\times l}$, where $F$ is the force, $I$ is the current and $l$ is the length.
The dimensional formula of force is $\text{ }\\!\\![\\!\\!\text{ ML}{{\text{T}}^{-2}}]$
The dimensional formula of current is $\text{ }\\!\\![\\!\\!\text{ A }\\!\\!]\\!\\!\text{ }$
The dimensional formula of length is $\text{ }\\!\\![\\!\\!\text{ L }\\!\\!]\\!\\!\text{ }$
Thus, the dimensional formula of magnetic field will be,
$\left[ \text{B} \right]\text{= }\dfrac{\left[ \text{ML}{{\text{T}}^{\text{-2}}} \right]}{\text{ }\\!\\![\\!\\!\text{ A }\\!\\!]\\!\\!\text{ }\\!\\!\times\\!\\!\text{ }\\!\\![\\!\\!\text{ L }\\!\\!]\\!\\!\text{ }}\text{= }\\!\\![\\!\\!\text{ M}{{\text{T}}^{\text{-2}}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }$
So, the dimensional formula of magnetic flux will be,
$\left[ \phi \right]\text{= }\\!\\![\\!\\!\text{ M}{{\text{T}}^{\text{-2}}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }\\!\\!\times\\!\\!\text{ }\\!\\![\\!\\!\text{ }{{\text{L}}^{\text{2}}}\text{ }\\!\\!]\\!\\!\text{ = }\\!\\![\\!\\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }$
Hence the dimensional formula of self-inductance will be,
$\Rightarrow \dfrac{\text{ }\\!\\![\\!\\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ A }\\!\\!]\\!\\!\text{ }}\text{= }\\!\\![\\!\\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}{{\text{A}}^{\text{-2}}}\text{ }\\!\\!]\\!\\!\text{ }$
So, the correct option is C.
Additional information: Henry (H) is the SI unit of inductance. Both self-inductance and mutual inductance use the Henry to represent the inductance.
$H=kg{{m}^{2}}{{s}^{-2}}{{A}^{-2}}$
One Henry can write like this also, one-kilogram meter squared per second square per ampere squared.
When a current change at the rate of 1 ampere per second and induced emf is one volt, the self-inductance of the coil will be one Henry.

We can find the self inductance of a solenoid.
Consider a solenoid of length l with n number of turns. If a current i flows through this solenoid, a magnetic field will generate inside.
Magnetic field, $B={{\mu }_{0}}\dfrac{Ni}{l}$
Each turn has area A, then the total magnetic flux through the solenoid is given by; ${{\phi }_{\mathbf{B}}}={{\mu }_{0}}\dfrac{Ni}{l}AN$
i.e. ${{\phi }_{\mathbf{B}}}={{\mu }_{0}}\dfrac{{{N}^{2}}i}{l}A$
Varying current will generate induced emf.
So, induced emf, $e=-\dfrac{d}{dt}\left[ {{\mu }_{0}}\dfrac{{{N}^{2}}i}{l}A \right]$
Negative sign indicates the generated emf is opposing the cause producing it.
Constant terms can be taken outside, $e=-{{\mu }_{0}}{{N}^{2}}\dfrac{A}{l}\dfrac{di}{dt}$
These constant terms are collectively treated as proportionality constant and known as self-inductance.
$e=-L\dfrac{di}{dt}$
Therefore, $L=\dfrac{{{\mu }_{0}}{{N}^{2}}A}{l}$
$L={{\mu }_{0}}{{n}^{2}}Al$, where n is the number of turns (N) per unit length (l)

Note: $F=qvB$, where charge can be written as the product of current and time. Here we used $F=B\times I\times l$. Both are the same. If the dimensional formula of force is not known, then we can reduce the force again to the fundamental form.