
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
126.9k+ views
Hint: 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.
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.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application

Young's Double Slit Experiment Step by Step Derivation

JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main
