Consider the equation H = frac{x^p epsilon^q | vedclass

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(A) The dimensional formulas are: $[H] = [M^0 L^0 T^0 A^1 L^{-1}] = [L^{-1} A^1]$, $[x] = [L^1]$, $[\epsilon] = [M^{-1} L^{-3} T^4 A^2]$, $[E] = [M^1

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$1, 1, 1, 1$

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(A) The dimensional formulas are: $[H] = [M^0 L^0 T^0 A^1 L^{-1}] = [L^{-1} A^1]$, $[x] = [L^1]$, $[\epsilon] = [M^{-1} L^{-3} T^4 A^2]$, $[E] = [M^1 L^1 T^{-3} A^{-1}]$, and $[t] = [T^1]$.Substituting these into the equation $H = x^p \epsilon^q E^r t^{-s}$:$[L^{-1} A^1] = [L]^p [M^{-1} L^{-3} T^4 A^2]^q [M^1 L^1 T^{-3} A^{-1}]^r [T]^{-s}$$[L^{-1} A^1] = M^{-q+r} L^{p-3q+r} T^{4q-3r-s} A^{2q-r}$Comparing the powers on both sides:For $M$: $-q + r = 0 \Rightarrow q = r$For $A$: $2q - r = 1 \Rightarrow 2q - q = 1 \Rightarrow q = 1, r = 1$For $L$: $p - 3q + r = -1 \Rightarrow p - 3(1) + 1 = -1 \Rightarrow p - 2 = -1 \Rightarrow p = 1$For $T$: $4q - 3r - s = 0 \Rightarrow 4(1) - 3(1) - s = 0 \Rightarrow 1 - s = 0 \Rightarrow s = 1$Thus, $p=1, q=1, r=1, s=1$. The correct option is $A$.

Consider the equation $H = \frac{x^p \epsilon^q E^r}{t^s}$, where $H = \text{magnetic field intensity}$, $E = \text{electric field}$, $\epsilon = \text{permittivity}$, $x = \text{distance}$, and $t = \text{time}$. The values of $p, q, r$, and $s$ respectively are:

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