Q163 / 180 · COMEDK 2019 · Physics
The dimensional formula for inductance is
- A$\left[\mathrm{ML}^{2} \mathrm{~T}^{-1} \mathrm{~A}^{-2}\right]$
- B$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]$
- Correct answerC$\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]$
- D$\left[\mathrm{ML}^{2} \mathrm{TA}^{-2}\right]$
Solution
EMF induced in circular coil of self-inductance $L$ is given by $\varepsilon=-L \frac{d I}{d t}$
$\mathrm{EMF}$ is equal to the potential difference dimensionally.
Hence,
$$
\varepsilon=V
$$
We know,
$$
V=\frac{W}{q}
$$
Thus, dimensional formula of potential difference is given as
$$
\begin{aligned}
V &=\frac{\left[M^{1} L^{2} T^{-2}\right]}{[A T]} \\
&=\left[M^{1} L^{2} T^{-3} A^{-1}\right]
\end{aligned}
$$
The self-inductance is given by, $L=V \times \frac{d t}{d I}$
Dimensional formula of self-inductance is given by,
$$
L=\left[\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right] \times \frac{[\mathrm{T}]}{[\mathrm{A}]}
$$
Dimensional formula of self-inductance,
$$
L=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]
$$
Hence, the correct answer is option: 'C'.