Explanation

$L \propto v^{x}A^{y}F^{z} \Rightarrow L = kv^{x}A^{y}F^{z}$ Putting the dimensions in the above relation $\left\lbrack ML^{2}T^{- 1} \right\rbrack = k\left\lbrack LT^{- 1} \right\rbrack^{x}\left\lbrack LT^{- 2} \right\rbrack^{y}\left\lbrack MLT^{- 2} \right\rbrack^{z}$ $\Rightarrow \left\lbrack ML^{2}T^{- 2} \right\rbrack = k\lbrack M^{z}L^{x + y + z}T^{- x - 2y - 2z}\rbrack$ Comparing the powers of $M,\ L$ and $T$ $z = 1$ …(i) $x + y + z = 2$ …(ii) $- x - 2y - 2z = - 1$ …(iii) On solving (i), (ii) and (iii) $x = 3,\ y = - 2,\ z = 1$ So dimension of $L$ in terms of $v,\ A$ and $f$ $\lbrack L\rbrack = \lbrack Fv^{3}A^{- 2}\rbrack$