JEE Mains Physics · Units and Measurements

Dimensions of Electric and Magnetic Quantities

Electric and magnetic quantities add current A as a fourth base quantity; each one's dimensions come from a defining equation, and a few combinations reduce to a speed, a resistance, a time or an energy density.

All Units and Measurements notes JEE Mains Physics strategy

Why this matters

Twenty-five PYQs, all multiple choice, and four from 2026. Eleven build dimensions from defining equations such as V = W/q or F = qvB, five of them as match-the-list; fourteen ask about a combination such as 1/(μ₀ε₀), ε₀E², E/H or R/√(LC). The combinations are quicker to recognise than to work out, which is why the second half of the page is a list of shortcuts.

Concept 1 of 2: Dimensions of electric and magnetic quantities

Current A is a base quantity, so charge is current × time, AT. Every other electric quantity comes from an equation that links it to energy or force: a voltage is energy per charge, a field is force per charge, a magnetic field comes from the force

qvB

on a moving charge. Build each one from that equation.

Definition

Charge $q = It$ : $AT$ .
Potential and emf $V = W/q$ : $ML^{2}T^{-3}A^{-1}$ . Resistance $R = V/I$ : $ML^{2}T^{-3}A^{-2}$ .
Capacitance $C = q/V$ : $M^{-1}L^{-2}T^{4}A^{2}$ . Inductance (self or mutual), from $U = \tfrac{1}{2}LI^{2}$ : $ML^{2}T^{-2}A^{-2}$ .
Electric field $E = F/q$ : $MLT^{-3}A^{-1}$ . Magnetic field, from $F = qvB$ : $MT^{-2}A^{-1}$ . Magnetic flux $BA$ : $ML^{2}T^{-2}A^{-1}$ .
$\varepsilon_0$ , from Coulomb's law: $M^{-1}L^{-3}T^{4}A^{2}$ . $\mu_0$ , from $B = \mu_0I/(2\pi r)$ : $MLT^{-2}A^{-2}$ .
Magnetic moment $IA$ : $L^{2}A$ . Magnetising field H and magnetisation: $L^{-1}A$ .

Quantity	Defining relation	Dimensions
Charge	$q = It$	$AT$
Potential difference, emf	$V = W/q$	$ML^{2}T^{-3}A^{-1}$
Resistance	$R = V/I$	$ML^{2}T^{-3}A^{-2}$
Resistivity	$\rho = RA/l$	$ML^{3}T^{-3}A^{-2}$
Capacitance	$C = q/V$	$M^{-1}L^{-2}T^{4}A^{2}$
Self or mutual inductance	$U = \tfrac{1}{2}LI^{2}$	$ML^{2}T^{-2}A^{-2}$ A⁻², not A⁻¹: energy divided by current squared.
Electric field	$E = F/q$	$MLT^{-3}A^{-1}$
Magnetic field (induction) B	$F = qvB$	$MT^{-2}A^{-1}$
Magnetic flux	$\Phi = BA$	$ML^{2}T^{-2}A^{-1}$
Permittivity $\varepsilon_0$	$F = q_1q_2/(4\pi\varepsilon_0r^{2})$	$M^{-1}L^{-3}T^{4}A^{2}$
Permeability $\mu_0$	$B = \mu_0I/(2\pi r)$	$MLT^{-2}A^{-2}$
Magnetic moment	$m = IA$	$L^{2}A$
Magnetising field H, magnetisation	$H = B/\mu_0$	$L^{-1}A$
Electric dipole moment	$p = qd$	$LTA$

Charge is AT; every other entry follows from its defining relation.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q1Moderate

Example 1 · Units and Measurements · Dimensions of Electric and Magnetic Quantities

Match the List-I with List-II

List-I	List-II
(A) Magnetic induction	(I) $MLT^{-2}A^{-2}$
(B) Magnetic flux	(II) $ML^{2}T^{-2}A^{-2}$
(C) Magnetic permeability	(III) $ML^{0}T^{-2}A^{-1}$
(D) Self inductance	(IV) $ML^{2}T^{-2}A^{-1}$

Choose the correct answer from the options given below:

Inductance has A⁻², flux has A⁻¹

Magnetic flux is ML²T⁻²A⁻¹ and inductance is flux per current, ML²T⁻²A⁻². The two entries differ only in the power of A, and a match list places them side by side.

Charge is AT, not a base quantity

The base quantity is current. When a question uses charge Q as a base instead, rewrite A as QT⁻¹: capacitance then becomes M⁻¹L⁻²T²Q².

Drill 10 more on dimensions of electric and magnetic quantities

Concept 2 of 2: Electromagnetic combinations with simple dimensions

Several combinations collapse to something familiar, and recognising them saves the whole calculation.

1/\sqrt{\mu_0\varepsilon_0}

is the speed of light.

\tfrac{1}{2}\varepsilon_0E^{2}

is the energy stored per volume in an electric field.

RC

L/R

and

\sqrt{LC}

are the time scales of circuits. Replace the combination by what it means, then finish the dimensions.

Definition

$\dfrac{1}{\mu_0\varepsilon_0} = c^{2}$ : $L^{2}T^{-2}$ .
$\sqrt{\mu_0/\varepsilon_0}$ is a resistance (about 377 Ω); so is $E/H$ .
$\tfrac{1}{2}\varepsilon_0E^{2}$ and $\dfrac{B^{2}}{2\mu_0}$ are energy densities: $ML^{-1}T^{-2}$ .
$E/B$ is a speed. $\varepsilon_0E$ is a surface charge density (it has the dimensions of $D$ ), so $\varepsilon_0E/t$ is a current density, A m⁻².
$RC$ , $L/R$ and $\sqrt{LC}$ are times. $L/C$ is a resistance squared, not a time.
$\dfrac{e^{2}}{4\pi\varepsilon_0hc}$ has no dimensions. $\mu_0$ and $\varepsilon_0$ DO have dimensions; relative permeability, power factor and quality factor do not.

Combinations to recognise

\frac{1}{\mu_0\varepsilon_0} = c^{2} \qquad \sqrt{\frac{\mu_0}{\varepsilon_0}} = [R] \qquad \tfrac{1}{2}\varepsilon_0E^{2} = \frac{B^{2}}{2\mu_0} = [\text{energy/volume}] \qquad [RC] = [L/R] = [\sqrt{LC}] = T

Worked example

Find the dimensions of

\varepsilon_0E^{2}c

, where E is an electric field and c is the speed of light. Which physical quantity has these dimensions?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q2Moderate

Example 2 · Units and Measurements · Dimensions of Electric and Magnetic Quantities

L, C and R represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula

ML^{2}{\text{ }T}^{- 4}{\text{ }A}^{- 2}

corresponds to

ε₀E² is energy per volume, not energy

½ε₀E² is the energy stored per unit volume, so its dimensions are ML⁻¹T⁻², the same as pressure. Writing ML²T⁻² misses the division by volume.

μ₀ is not dimensionless

μ₀ has dimensions MLT⁻²A⁻². The dimensionless ones are ratios: relative permeability μ/μ₀, dielectric constant, power factor cos φ and the quality factor.

Drill 13 more on electromagnetic combinations with simple dimensions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Electromagnetic combinations with simple dimensions
Combinations to recognise
$\frac{1}{\mu_0\varepsilon_0} = c^{2} \qquad \sqrt{\frac{\mu_0}{\varepsilon_0}} = [R] \qquad \tfrac{1}{2}\varepsilon_0E^{2} = \frac{B^{2}}{2\mu_0} = [\text{energy/volume}] \qquad [RC] = [L/R] = [\sqrt{LC}] = T$

Reference tables (1)

Dimensions of electric and magnetic quantities14 rows

Quantity	Defining relation	Dimensions
Charge	$q = It$	$AT$
Potential difference, emf	$V = W/q$	$ML^{2}T^{-3}A^{-1}$
Resistance	$R = V/I$	$ML^{2}T^{-3}A^{-2}$
Resistivity	$\rho = RA/l$	$ML^{3}T^{-3}A^{-2}$
Capacitance	$C = q/V$	$M^{-1}L^{-2}T^{4}A^{2}$
Self or mutual inductance	$U = \tfrac{1}{2}LI^{2}$	$ML^{2}T^{-2}A^{-2}$ A⁻², not A⁻¹: energy divided by current squared.
Electric field	$E = F/q$	$MLT^{-3}A^{-1}$
Magnetic field (induction) B	$F = qvB$	$MT^{-2}A^{-1}$
Magnetic flux	$\Phi = BA$	$ML^{2}T^{-2}A^{-1}$
Permittivity $\varepsilon_0$	$F = q_1q_2/(4\pi\varepsilon_0r^{2})$	$M^{-1}L^{-3}T^{4}A^{2}$
Permeability $\mu_0$	$B = \mu_0I/(2\pi r)$	$MLT^{-2}A^{-2}$
Magnetic moment	$m = IA$	$L^{2}A$
Magnetising field H, magnetisation	$H = B/\mu_0$	$L^{-1}A$
Electric dipole moment	$p = qd$	$LTA$

Charge is AT; every other entry follows from its defining relation.

Watch out for (4)

Inductance has A⁻², flux has A⁻¹
→ Dimensions of electric and magnetic quantities
Charge is AT, not a base quantity
→ Dimensions of electric and magnetic quantities
ε₀E² is energy per volume, not energy
→ Electromagnetic combinations with simple dimensions
μ₀ is not dimensionless
→ Electromagnetic combinations with simple dimensions

Test yourself on Units and Measurements

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Units and Measurements chapter test Drill all 25 questions