Q97 / 180 · AIIMS 2003 · Physics

Using mass $(M)$, length $(L)$, time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permeability is

  • A
    $\mathrm{M}^{-1} \mathrm{LT}^{-2} \mathrm{~A}$
  • B
    $\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}$
  • Correct answer
    C
    $\mathrm{MLT}^{-2} \mathrm{~A}^{-2}$
  • D
    $\mathrm{MLT}^{-1} \mathrm{~A}^{-1}$.

Solution

We know that the force per unit length of a wire carrying current due to another parallel wire carrying current is given by $$ \begin{gathered} \frac{d F}{d l}=\frac{\mu_0 i_1 i_2}{2 \pi d} \Rightarrow \mu_0=\frac{2 \pi d}{i_1 i_2} \frac{d F}{d l} \\ \therefore \quad\left[\mu_0\right]=\frac{[\mathrm{L}]}{\left[\mathrm{A}^2\right]} \cdot \frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]} \\ \text { or }\left[\mu_0\right]=\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{A}^{-2}\right]=\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right] . \end{gathered} $$

Hence, the correct answer is option: 'C'.

Relevant Questions