# How to find the dimension formula for inductance and also the dimension for resistance ?

Sep 8, 2017

Dimensions of L, $M {T}^{- 2} {L}^{2} {A}^{- 2}$

Dimensions of R,

$M {L}^{2} {T}^{- 3} {A}^{- 2}$

#### Explanation:

Firstly consider resistance.

It's defining equation is, Ohm's law,

$V = I R$
$\implies R = \frac{V}{I}$

Now $V$ has units of (electric field)*(distance).

But electric field has units (force)/(charge).

Also, charge has dimensions of (current)(time) and force has dimensions (mass)(length)/(time)^2.

Thus, dimensions of $V$ is,

$\left[V\right] = \frac{L M L {T}^{- 2}}{A T}$
$\implies \left[V\right] = M {L}^{2} {T}^{- 3} {A}^{- 1}$

Current $I$ has dimensions $\left[I\right] = A$

Thus, dimensions of resistance,

$\left[R\right] = \frac{\left[V\right]}{\left[I\right]} = M {L}^{2} {T}^{- 3} {A}^{- 2}$

For inductance, the defining equation is,

$\phi = L I$

But $\phi$ has units (magnetic field)*(length)^2

Magnetic field from Lorentz force law has units, (Force)(velocity)^(-1)(charge)^(-1)

Therefore, dimensions of magnetic field,

$\left[B\right] = \frac{M L {T}^{- 2}}{L {T}^{- 1} A T}$
$\implies \left[B\right] = \frac{M L {T}^{- 2}}{L A}$
$\implies \left[B\right] = M {T}^{- 2} {A}^{- 1}$

Therefore dimensions of magnetic flux,

$\left[\phi\right] = \left[B\right] {L}^{2}$
$\implies \left[\phi\right] = M {T}^{- 2} {L}^{2} {A}^{- 1}$

Thus finally, dimensions of inductance,

$\left[L\right] = \frac{\left[\phi\right]}{\left[I\right]}$
$\implies \left[L\right] = M {T}^{- 2} {L}^{2} {A}^{- 2}$