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Q. 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 $k=\frac{1}{4 \pi \epsilon_0}$, where $\epsilon_0$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B]=[e]^\alpha\left[m_e\right]^\beta[h]^\gamma[k]^\delta$. The value of $\alpha+\beta+\gamma+\delta$ is _____

JEE AdvancedJEE Advanced 2022

Solution:

$ B = e ^\alpha\left( m _{ e }\right)^\beta h ^\gamma k ^\delta$
${[ B ]=\left[ e ^\alpha\right]\left[ m _{ e }\right]^\beta[ h ]^\gamma\left[ k ^\delta\right]} $
$ {\left[ M ^1 T ^{-2} A ^{-1}\right]=[ AT ]^\alpha\left[ m ^\beta\right]^{[}\left[ ML ^2 T ^{-1}\right]^\gamma\left[ ML ^3 A ^{-2} T ^{-4}\right]^\delta} $
$M ^1 T ^{-2} A ^{-1}= m ^{\beta+\gamma+\delta} L ^{2 r +3\delta} T ^{\alpha-\gamma-4\delta} A ^{\alpha-2 \delta}$
Compare : $\beta+\gamma+\delta=1 ; 2 \gamma+3 \delta=0, \alpha-\gamma-4 \delta=-2, \alpha-2 \delta=-1$
On solving $\alpha=3, \beta=2, \gamma=-3, \delta=2$
$\alpha+\beta+\gamma+\delta=4$