Consider the equation H = (xp εq Er)/(ts) Where H = magnetic field intensity; E = electric field, ε = permittivity, x = distance, t = time The values of p, q, r and s respectively are: (a) 1, 1, 1, 1 (b) -1, 1, 2, 1 (c) 1, -1, -2, 1 (d) -1, -2, -2, 1

Answer: (a). [H] = [A L⁻¹], [ε] = [M⁻¹L⁻³T⁴A²], [E] = [MLT⁻³A⁻¹]. Try p=q=r=s=1: [xε E/t] = [L][M⁻¹L⁻³T⁴A²][MLT⁻³A⁻¹][T⁻¹]. = [M⁰ L⁻¹ T⁰ A], which equals [H]. So the answer is (a) 1, 1, 1, 1.

Consider the equation H = (xp εq Er)/(ts) Where H = magnetic field… | JEE Main 2026 (8 Apr, Shift 2)

Q1JEE Main 2026 · 8 APR · SHIFT 2Easy

Applications of Dimensional Analysis

Consider the equation $H = \frac{x^p \epsilon^q E^r}{t^s}$

Where $H =$ magnetic field intensity; $E =$ electric field, $\epsilon =$ permittivity, $x =$ distance, $t =$ time The values of $p, q, r$ and $s$ respectively are:

Show answer and solution

Answer: (a)

$[H] = [A L^{-1}]$ , $[\epsilon] = [M^{-1}L^{-3}T^4A^2]$ , $[E] = [MLT^{-3}A^{-1}]$ .
Try $p=q=r=s=1$ : $[x\epsilon E/t] = [L][M^{-1}L^{-3}T^4A^2][MLT^{-3}A^{-1}][T^{-1}]$ .
$= [M^0 L^{-1} T^0 A]$ , which equals $[H]$ .
So the answer is (a) $1, 1, 1, 1$ .

Report a mistake

Applications of Dimensional Analysis question from JEE Main 2026 (8 Apr, Shift 2)

More questions on Applications of Dimensional Analysis