$X = 3YZ^2$, find the dimension of $Y$ in (MKSA) system, if $X$ and $Z$ are the dimensions of capacity and mag

Question

$X = 3YZ^2$, find the dimension of $Y$ in (MKSA) system, if $X$ and $Z$ are the dimensions of capacity and magnetic field respectively.

Accepted Answer

1. **Find the dimensions of capacity ($X$):** Capacitance $C = \frac{Q}{V} = \frac{Q^2}{W}$.
Since $Q = [AT]$ and Work $W = [ML^2T^{-2}]$, 
$[X] = \frac{[AT]^2}{[ML^2T^{-2}]} = [M^{-1} L^{-2} T^4 A^2]$.

2. **Find the dimensions of magnetic field ($Z$):** Magnetic force $F = qvB \implies B = \frac{F}{qv}$.
Since $F = [MLT^{-2}]$, $q = [AT]$, and $v = [LT^{-1}]$,
$[Z] = \frac{[MLT^{-2}]}{[AT][LT^{-1}]} = [M T^{-2} A^{-1}]$.

3. **Calculate the dimensions of $Y$:** 
Given $X = 3YZ^2 \implies Y = \frac{X}{3Z^2}$. Ignoring the dimensionless constant $3$, we get:
$[Y] = \frac{[X]}{[Z]^2} = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M T^{-2} A^{-1}]^2}$
$[Y] = \frac{[M^{-1} L^{-2} T^4 A^2]}{[M^2 T^{-4} A^{-2}]} = [M^{-3} L^{-2} T^8 A^4]$.

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