Question

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

Answer 1

Dimensions of L, `MT^(-2)L^2A^(-2)`

Dimensions of R,

`ML^2T^(-3)A^(-2)`

Explanation:

Firstly consider resistance.

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

`V = IR`
`implies R = 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,

`[V] = (LMLT^(-2))/(AT)`
`implies [V] = ML^2T^(-3)A^(-1)`

Current `I` has dimensions `[I] =A`

Thus, dimensions of resistance,

`[R] = [[V]]/[[I]] = ML^2T^(-3)A^(-2)`

For inductance, the defining equation is,

`phi = LI`

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,

`[B] = (MLT^(-2))/(LT^(-1)AT)`
`implies [B] = (MLT^(-2))/(LA)`
`implies [B] = MT^(-2)A^(-1)`

Therefore dimensions of magnetic flux,

`[phi] = [B]L^2`
`implies [phi] = MT^(-2)L^2A^(-1)`

Thus finally, dimensions of inductance,

`[L] = [[phi]]/[[I]]`
`implies [L] = MT^(-2)L^2A^(-2)`