User talk:Muneeb muzzamil

Ohm's law and drift velocity in conductors From Physclips: Mechanics with animations and film.

This page relates the DC electric current to the drift velocity of charge carriers, and thus shows how it depends on their number density, their charge, the cross sectional area of the conductor, the electric field and properties of the material. We begin by introducing an analogy using mass and gravitational field instead of charge and electric field. If you prefer, go straight to the electrical case.

How much rain falls in a bucket per second? The top of a bucket has area A. Raindrops of mass m are falling at a constant speed v (their terminal velocity). There are n drops per m3. Let's look at the cylinder (length L, area A) above the bucket in the diagram at right. It contains a number of raindrops given by

n*volume = nAL.

A drop at the very top of the cylinder will leave the cylinder after a time t = L/v. So the time t for all of the raindrops in the cylinder to leave is

t = L/v

So the rate of rainfall (in kilograms of water per second) into the bucket is

I = mass/time = (nALm)/(L/v) = nAvm. (1)

Now let's look at the terminal velocity of the rain drops, which we have assumed constant. The weight pulls them down, and air resistance or drag acts in the upwards direction. If they start from rest, there would be no drag, and their weight would accelerate them downwards until they reached their terminal velocity v. This is the velocity at which there is no further acceleration, because the total force is zero, ie

weight = drag force.

sketch of rain in bucket

For small droplets, the drag force is proportional to their speed. It also depends upon their radius, and on the viscosity of the air, but for raindrops of fixed size, let's include these effects in a constant of proportionality Kdrag. We write:

Kdragv = Fdrag = weight = mg.

We use this to substitute for v in equation (1) and get:

I = mass/time = nAvm = nAm2g/Kdrag

Here we note that the current is:

proportional to the number of mass carriers/cubic metre of air (n) proportional to the cross sectional area of 'conductor' (A) proportional to the square of the mass on an individual mass carrier (m) proportional to the magnitude of the field that moves them (g) inversely proportional to the proportionality constant for drag (Kdrag)

(The proportionality to mass squared should be qualified: Kdrag is only constant if the size is unchanged, so this proportionality refers to changing the mass by changing the density of the material from which the drops are made.) Let's now derive Ohm's law for gravity and raindrops. We'll need the gravitational potential Vgrav, which is the gravitational potential energy per unit mass, exactly analogous to the electrical case. Take zero potential energy at height h = 0 and we get:

Vgrav = Ugrav/m = mgh/m = gh.

In the diagram, the cylinder is our resistor (its air is resisting the motion of raindrops). From one end to the other, the height changes by L, so the (gravitational) potential difference across it is Vgrav = gL. So, to get the gravitational version of Ohm's law:

Vgrav/I = (gL)Kdrag/nAm2g = = LKdrag/nAm2.

Let's separate out the geometrical parameters L and A, and combine the other constants (Kdrag, n and m) into one new constant for this particular air+rain mix:

Vgrav/I = (Kdrag/nm2).(L/A) = ρ.(L/A),      where ρ = Kdrag/nm2

is the resisitivity of this particular air+rain mix. (In terms of the analogy, this mix is analogous to a particular substance that conducts electricity.) Now let's think about this particular cylinder, with its given length and cross section (analogous to a particular piece of wire).

Vgrav/I = R      where R = ρL/A,

and where R is the resistance of this particular cylinder of this particular 'substanc