# Deriving V(t) = L (di/dt)

1. Mar 20, 2013

### CraigH

Voltage across an inductor at any moment in time can be calculated as the inductance multiplied by the rate of change of current.

How is this equation derived?
I'm pretty sure it comes from Faraday law

-emf = rate of change of magnetic flux

but I cannot find the relationship.

Thanks!

2. Mar 20, 2013

### phyzguy

You've pretty much hit on it. Since the EMF (voltage) = d/dt(magnetic flux), and the magnetic flux is proportional to the current, then the voltage is proportional to d/dt(current). The proportionality constant is the inductance. You could consider this as the definition of inductance.

3. Mar 20, 2013

### CraigH

Okay, so does magnetic flux equal current multiplied by inductance?

The units check out:
base units
M = mass
T = time
Q = charge
L = length

magnetic flux = M(T^-1)(Q^-1)(L^2)
current = Q(T^-1)
inductance = M(L^2)(Q^-2)

M(T^-1)(Q^-1)(L^2) = Q(T^-1) M(L^2)(Q^-2)

4. Mar 21, 2013

### Emilyjoint

Yes.
Flux linkage = L x I
This means that an inductance of 1 Henry will produce a flux linkage of 1 weber for a current of 1 amp.
In terms of induced emf it means that an inductance of 1 Henry will generate an emf of 1 volt when the current changes at a rate of 1 amp per second.

Last edited: Mar 21, 2013
5. Mar 21, 2013

### Philip Wood

If there are no ferromagnetic materials present, then $n\Phi = LI$ and $\frac{dn\Phi}{dt} = L\frac{dI}{dt}$ are completely equivalent, and either can be used to define L. [Notation: $n\Phi$ = flux linkage.]

If there are ferromagnetic materials - as is often the case for inductors of large inductance - then the flux linkage is not strictly proportional to the current, that is L in the first equation is not a constant, so the second equation does not follow from the first by differentiation. If we're not too fussy, and the currents are small enough for the ferromagnetic material not to enter the 'saturation' region, we can take L in the first equation as approximately constant and stop worrying!

6. Mar 21, 2013

### CraigH

Brilliant answer! Thank you this has helped a lot.

7. Jun 24, 2016