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In a particular system of units, a physical quantity can be expressed in terms of the electric charge $c$, electron mass $m_c$, 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 $[B]=[c]^\alpha\left[m_c\right]^\beta[h]^\gamma[k]^\delta$. The value of $\alpha+\beta+\gamma+\delta$ is. . . . .
$3$
$4$
$5$
$6$
Solution
$B = e ^\alpha\left( m _e\right)^\beta h ^\gamma k ^\delta$
${[ B ]=\left[ e ^\alpha\right]\left[ m _e\right]^\beta[ h ]^\gamma\left[ k ^\delta\right]}$
${\left[ M ^1 T ^{-2} A ^{-1}\right]=[ AT ]^\alpha[ m ]^\beta\left[ ML ^2 T ^{-1}\right]^\gamma\left[ ML ^3 A ^{-2} T ^{-4}\right]^\delta}$
$M ^1 T ^{-2} A ^{-1}= m ^{\beta+\gamma+\delta} L ^{2 x +38} T ^{\alpha-\gamma-\alpha \delta} A ^{\alpha-28}$
Compare : $\beta+\gamma+\delta=1 ; 2 \gamma+3 \delta=0, \alpha-\gamma-4 \delta=-2, \alpha-2 \delta=-1$
On solving $\alpha=3, \beta=2, \gamma=-3, \delta=2$
$\alpha+\beta+\gamma+\delta=4$
