Question:
Which of the following combinations has the dimension of electrical resistance (\(\in_0\) is the permittivity of vacuum and \(\mu_0\) is the permeability of vacuum)?
Which of the following combinations has the dimension of electrical resistance (\(\in_0\) is the permittivity of vacuum and \(\mu_0\) is the permeability of vacuum)?
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\(\sqrt{\mu_0/\varepsilon_0}\) is the impedance of free space (about 377 ohm).
Updated On: Nov 5, 2026
- \(\sqrt{\dfrac{\in_0}{\mu_0}}\)
- \(\dfrac{\in_0}{\mu_0}\)
- \(\sqrt{\dfrac{\mu_0}{\in_0}}\)
- \(\dfrac{\mu_0}{\in_0}\)
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The Correct Option is C
Solution and Explanation
Dimensions of resistance: \([R] = \dfrac{[V]}{[I]} = \dfrac{ML^2T^{-3}A^{-1}}{A} = ML^2T^{-3}A^{-2}\).
\([\mu_0]\): from \(F = \dfrac{\mu_0 I_1 I_2 \ell}{2\pi d}\), \([\mu_0] = MLT^{-2}A^{-2}\).
\([\in_0]\): from \(F = \dfrac{q^2}{4\pi \in_0 r^2}\), \([\in_0] = M^{-1}L^{-3}T^{4}A^{2}\).
\(\dfrac{\mu_0}{\in_0} = \dfrac{MLT^{-2}A^{-2}}{M^{-1}L^{-3}T^{4}A^{2}} = M^2L^4T^{-6}A^{-4}\).
Therefore \(\sqrt{\dfrac{\mu_0}{\in_0}} = ML^2T^{-3}A^{-2}\), which equals \([R]\).
Answer: (3) \(\sqrt{\mu_0/\in_0}\)
\([\mu_0]\): from \(F = \dfrac{\mu_0 I_1 I_2 \ell}{2\pi d}\), \([\mu_0] = MLT^{-2}A^{-2}\).
\([\in_0]\): from \(F = \dfrac{q^2}{4\pi \in_0 r^2}\), \([\in_0] = M^{-1}L^{-3}T^{4}A^{2}\).
\(\dfrac{\mu_0}{\in_0} = \dfrac{MLT^{-2}A^{-2}}{M^{-1}L^{-3}T^{4}A^{2}} = M^2L^4T^{-6}A^{-4}\).
Therefore \(\sqrt{\dfrac{\mu_0}{\in_0}} = ML^2T^{-3}A^{-2}\), which equals \([R]\).
Answer: (3) \(\sqrt{\mu_0/\in_0}\)
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