At room temperature, if the relative permittivity of water be 80 and the velocity of light in water is 2.25 x 108 ms-1, then what is the relative permeability of water ?
Answer & explanation
Correct answer: option 3
Expressions:
μ = = \(\frac{1}{v^{2}ε}\)
c = 1/\(\sqrt { \mu_{o} }\epsilon_{o} \)
μr = μ/μo ⇒ μ = μrμo
εr = ε/εo ⇒ ε = εrεo
εr = 80 ; v = 2.25 x 108 ms-1 ; c = 3.0 x 108 ms-1
μr = 0.022