Question

In the formula $:X=3Y{Z}^2,X$ and $Z$ have dimensions of capacitance and magnetic induction respectively. What are the dimension of $Y$ in $MKSQ$ system?

Answer 1

The given relation is:

$X=3Y{Z}^2$

$Y=\frac{1}{3}\frac{X}{Z^2}$

since $X$ is the capacitance. So, its dimensions are:

$X=M^{-1}{L}^{-2}{T}^2{Q}^2$

and $Z$ is the magnetic induction and its dimensions are

$Z=M{T}^{-1}{Q}^{-1}$

Therefore, $Y$ can be expressed as:

$Y=\frac{M^{-1}{L}^{-2}{T}^2{Q}^2}{{\left[M{T}^{-1}{Q}^{-1}\right]}^2}=M^{-3}{L}^{-2}{T}^4{Q}^4$