JEE Advanced 2022 Paper 2 · Q04 · Dimensional Analysis

Question

In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$, electron mass $m_e$, Planck's constant $h$, and Coulomb's constant $k = \dfrac{1}{4\pi\epsilon_0}$, where $\epsilon_0$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B] = [e]^\alpha [m_e]^\beta [h]^\gamma [k]^\delta$. The value of $\alpha + \beta + \gamma + \delta$ is _____ .

SolveKar AI · Accepted Answer

Use SI dimensions: $[B] = M T^{-2} I^{-1}$, $[e] = I T$, $[m_e] = M$, $[h] = M L^2 T^{-1}$, $[k] = M L^3 T^{-4} I^{-2}$. Match exponents of $M, L, T, I$:
$M: 1 = \beta + \gamma + \delta$
$L: 0 = 2\gamma + 3\delta$
$T: -2 = \alpha - \gamma - 4\delta$
$I: -1 = \alpha - 2\delta$
From the $L$-equation $\gamma = -3\delta/2$; from the $I$-equation $\alpha = 2\delta - 1$. Substituting in the $T$-equation: $-2 = (2\delta-1) + 3\delta/2 - 4\delta = -\delta/2 - 1$, so $\delta = 2$. Then $\alpha = 3$, $\gamma = -3$, and from the $M$-equation $\beta = 1 - \gamma - \delta = 1 + 3 - 2 = 2$. Sum: $3 + 2 - 3 + 2 = 4$.

JEE Advanced 2022 Paper 2 · Q04 · Dimensional Analysis

Solution