If the dimension of a physical quantity EBμ is MaLbTc. Find the value of b−ca. HereE is magnitude of electric field B is Magnitude of magnetic field μ is Permeability of medium

EBμ$$=B2C$$μ $$=$$ energy density × $$speed=$$ Energy per unit volume × $$speed=Energyvolume$$×speed $$ML2T$$−$$2L3$$×LT−$$1=ML0T$$−$$3$$ Here $$a=1$$, $$b=0$$, $$c=$$−$$3$$ ⇒b−$$ca=0$$−$$($$−$$3)1=3$$

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If the dimension of a physical quantity $\frac{E B}{μ}$ is $M^{a} L^{b} T^{c}$ . Find the value of $(\frac{b - c}{a})$ . Here
E is magnitude of electric field
B is Magnitude of magnetic field
$μ$ is Permeability of medium

Moderate

Solution

$\frac{E B}{μ} = \frac{B^{2} C}{μ}$ = energy density $\times$ speed
= Energy per unit volume $\times$ speed
$= \frac{E n e r g y}{v o l u m e} \times s p e e d$
$\frac{M L^{2} T^{- 2}}{L^{3}} \times L T^{- 1} = M L^{0} T^{- 3}$
Here $a = 1, b = 0, c = - 3$
$\Rightarrow \frac{b - c}{a} = \frac{0 - (- 3)}{1} = 3$

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