The dimensional formula of {₀/ε₀ is

Correct Answer is : [ML2T−3A−2][M L² T^{-3} A^{-2}][ML2T−3A−2]

Correct Answer is : [ML2T−3A−2][M L² T^{-3} A^{-2}][ML2T−3A−2]

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BITSAT 2019 Slot 2 Solved Paper

Section: Physics

Question : 20 of 150

Marks: +1, -0

\sqrt{\frac{\mu_0}{\varepsilon_0}}

Solution:

We know that, velocity of light

c

is terms of

\varepsilon_0

and

\mu_0

is given by

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \Rightarrow c^{2} = \frac{1}{\mu_0 \varepsilon_0}

c^{2} = \frac{1}{\mu_0^{2}} \cdot \frac{\mu_0}{\varepsilon_0} \Rightarrow c = \frac{1}{\mu_0} \sqrt{\frac{\mu_0}{\varepsilon_0}}

\Rightarrow \sqrt{\frac{\mu_0}{\varepsilon_0}} = c \mu_0

.......(i)

Dimension of magnetic field,

B = \frac{\tau}{N I A} = \frac{[M L^2 T^{-2}]}{[A][L^2]}

= [M A^{-1} T^{-2}]

Again,

B = \mu_0 n I

\therefore \mu_0 = \frac{B}{n l} = \frac{[M A^{-1} T^{-2}]}{[L^{-1}][A]}

= [M L A^{-2} T^{-2}]

\mu_0 = [M L A^{-2} T^{-2}]

.......(ii)

and

C = [L T^{-1}]

.......(iii)

From Eqs. (i), (ii) and (iii), we get

\sqrt{\frac{\mu_0}{\varepsilon_0}} = [L T^{-1}][M L A^{-2} T^{-2}]

= [M L^2 T^{-3} A^{-2}]

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