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# Mean drift velocity watch

1. I need some help with the following question:

A copper rod is stretched in a tensile testing machine so that its diameter at the widest section is three times that at the narrowest section, the neck. The road is connected to an electrical battery

a) Calculate the ratio of the maximum to minimum mean drift velocities of the free electrons in the rod.

I don't really understand how to do this.

Should I try work out the area for each one and see what difference it makes idek. Can I just plug a random number for the diameter and then show it is that many times bigger?

Any help would be appreciated
2. (Original post by Super199)
I need some help with the following question:

A copper rod is stretched in a tensile testing machine so that its diameter at the widest section is three times that at the narrowest section, the neck. The road is connected to an electrical battery

a) Calculate the ratio of the maximum to minimum mean drift velocities of the free electrons in the rod.

I don't really understand how to do this.

Should I try work out the area for each one and see what difference it makes idek. Can I just plug a random number for the diameter and then show it is that many times bigger?

Any help would be appreciated
yeah get the ratio of areas... plugging in a convenient number would work though it'd be a bit more stylish to do it by algebra.

I = vena (whatever order you have learnt it or whatever order it appears in the formulae booklet)

Where I is the current, v is the mean drift speed of the electrons, n is the mean electron density, and a is the cross sectional area.

There is two cases - where the diameter is the maximum and where the diameter is the minimum. In both regions the current would have to be the same (Imagine water flowing through a thick pipe and then to a thin one - the volume of water flowing through any cross-section has to be the same) so you can equate both the vena. Notice that e and n are constants. So that boils down to:

Vmax * Amin = Vmin * Amax

The reason why V is max when A is min and vice versa is that V is inversely proportional to A. (The smaller the tube the faster the water molecules has to flow through).

Express A in terms of d, cancel out any common coeffients/constants and then find the ratio. This will be easy after this stage.

Hope it helps
4. (Original post by RoyalBlue7)

I = vena (whatever order you have learnt it or whatever order it appears in the formulae booklet)

Where I is the current, v is the mean drift speed of the electrons, n is the mean electron density, and a is the cross sectional area.

There is two cases - where the diameter is the maximum and where the diameter is the minimum. In both regions the current would have to be the same (Imagine water flowing through a thick pipe and then to a thin one - the volume of water flowing through any cross-section has to be the same) so you can equate both the vena. Notice that e and n are constants. So that boils down to:

Vmax * Amin = Vmin * Amax

The reason why V is max when A is min and vice versa is that V is inversely proportional to A. (The smaller the tube the faster the water molecules has to flow through).

Express A in terms of d, cancel out any common coeffients/constants and then find the ratio. This will be easy after this stage.

Hope it helps
Wow thanks for this. I'll give it a go in a min. Aren't you the guy that like got A* for every subject you took ?
5. (Original post by Super199)
Wow thanks for this. I'll give it a go in a min. Aren't you the guy that like got A* for every subject you took ?
You're welcome.

Yes...
6. Can you explain that further please? Didn't really get that..
7. (Original post by UnknownDude)
Can you explain that further please? Didn't really get that..
Charge is constant, because that's one of our series circuit rules.

As is charge carrier density, because the same material is being used. And the charge on an electron is constant regardless.

So, we have:

I=nave

Where only a and v are variables.

We can therefore rewrite this as

v=k/a

Or that mean drift velocity is inversely proportional to cross sectional area. This leads immediately to the result that the ration of the drift velocities is the inverse of the ration of the areas.

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