In terms of potential difference V,electric c | vedclass

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(C) We know that the energy density in an electric field is $u_E = \frac{1}{2} \varepsilon_0 E^2$ and in a magnetic field is $u_B = \frac{1}{2} \frac{

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$(A, C)$

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(C) We know that the energy density in an electric field is $u_E = \frac{1}{2} \varepsilon_0 E^2$ and in a magnetic field is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.Since $u_E = u_B$,we have $\varepsilon_0 E^2 = \frac{B^2}{\mu_0}$,which implies $\frac{E^2}{B^2} = \frac{1}{\mu_0 \varepsilon_0} = c^2$.Using $V = Ed$ and $B = \frac{\mu_0 I}{2\pi r}$,we can relate these quantities.Alternatively,using dimensional analysis:$[\varepsilon_0] = M^{-1} L^{-3} T^4 A^2$,$[\mu_0] = M L T^{-2} A^{-2}$,$[V] = M L^2 T^{-3} A^{-1}$,$[I] = A$,$[c] = L T^{-1}$.For $(A)$: $[\mu_0 I^2] = (M L T^{-2} A^{-2})(A^2) = M L T^{-2}$. $[\varepsilon_0 V^2] = (M^{-1} L^{-3} T^4 A^2)(M^2 L^4 T^{-6} A^{-2}) = M L T^{-2}$. Thus,$(A)$ is dimensionally correct.For $(C)$: $[I] = A$. $[\varepsilon_0 c V] = (M^{-1} L^{-3} T^4 A^2)(L T^{-1})(M L^2 T^{-3} A^{-1}) = A$. Thus,$(C)$ is dimensionally correct.Therefore,the correct options are $(A)$ and $(C)$.

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In terms of potential difference $V$,electric current $I$,permittivity $\varepsilon_0$,permeability $\mu_0$,and speed of light $c$,the dimensionally correct equation$(s)$ is(are):$(A)$ $\mu_0 I^2 = \varepsilon_0 V^2$$(B)$ $\varepsilon_0 I = \mu_0 V$$(C)$ $I = \varepsilon_0 cV$$(D)$ $\mu_0 cI = V$