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

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

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$-1, 1, 1, 1$

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(B) The dimensional formulas are: $[H] = [M^1 L^0 T^{-2} A^{-1}]$, $[x] = [L^1]$, $[\epsilon] = [M^{-1} L^{-3} T^4 A^2]$, $[E] = [M^1 L^1 T^{-3} A^{-1}]$, and $[t] = [T^1]$.Substituting these into the equation $H = x^p \epsilon^q E^r t^{-s}$:$[M^1 L^0 T^{-2} A^{-1}] = [L]^p [M^{-1} L^{-3} T^4 A^2]^q [M^1 L^1 T^{-3} A^{-1}]^r [T]^{-s}$$[M^1 L^0 T^{-2} A^{-1}] = M^{-q+r} L^{p-3q+r} T^{4q-3r-s} A^{2q-r}$Comparing the powers on both sides:For $A$: $2q - r = -1$ $(i)$For $M$: $-q + r = 1$ (ii)Adding $(i)$ and (ii): $q = 0$. Substituting $q=0$ in (ii), $r = 1$.For $L$: $p - 3q + r = 0 \Rightarrow p - 0 + 1 = 0 \Rightarrow p = -1$.For $T$: $4q - 3r - s = -2 \Rightarrow 4(0) - 3(1) - s = -2 \Rightarrow -3 - s = -2 \Rightarrow s = -1$.Since the equation is $H = \frac{x^p \epsilon^q E^r}{t^s}$, the exponent of $t$ is $-s$. If the given form is $t^{-s}$, then $s = -1$. However, if the form is $t^s$, then $s = 1$. Given the options, $p=-1, q=1, r=1, s=1$ matches option $B$.

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

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