Calculating the number of turns of a solenoid.

[Asmaa Mohammad]

Homework Statement

Homework Equations

B = μ Ni/l

The Attempt at a Solution

Question no. a:[/B]

And Q no. b:

I am not sure about the step in the red circle in the first figure, and my book doesn't give a solution, so what is your opinion?

 

[cnh1995]

I am not sure about the step in the red circle in the first figure,

Looks good to me.

 

[Asmaa Mohammad]

Looks good to me.

I thought it is used in the case of a circular coil, but in the problem we deal with a solenoid, wouldn't this make a difference?

 

[cnh1995]

I thought it is used in the case of a circular coil, but in the problem we deal with a solenoid, wouldn't this make a difference?

Actually, it is appropriate for solenoid only. I don't see how you can use it for circular coil with constant radius.

 

[Asmaa Mohammad]

Actually, it is appropriate for solenoid only. I don't see how you can use it for circular coil with constant radius.

Suppose you have a wire with a length (l) and you want to make a circular coil with a radius (r), then the number of turns you get will be (N), and you can determine this N as this:
N = l / 2πr

 

[Merlin3189]

I thought it is used in the case of a circular coil, but in the problem we deal with a solenoid, wouldn't this make a difference?

I see what you mean: it's a helix rather than a circle. But the difference here is negligible, I think.

If the circumference of the solenoid is 20cm, and the pitch (the diameter of the wire) is 0.07356cm, then the length of wire per turn is sqrt(20^2 +0.07356^2)cm = 20.000135 cm giving 249.998 turns from 50m of wire.

But you are correct that in some cases with thick wire or not close wound, this could make a difference.

cnh1995 also makes an interesting point, that you can't wind a coil with several turns and have them all in exactly the same place. The best you might manage is a series of concentric short solenoids each with a slightly larger diameter. I can't agree that the standard formula is exactly correct in either situation, but in practical terms it is probably pretty close in most cases.

 

[Asmaa Mohammad]

I see what you mean: it's a helix rather than a circle. But the difference here is negligible, I think.

If the circumference of the solenoid is 20cm, and the pitch (the diameter of the wire) is 0.07356cm, then the length of wire per turn is sqrt(20^2 +0.07356^2)cm = 20.000135 cm giving 249.998 turns from 50m of wire.

But you are correct that in some cases with thick wire or not close wound, this could make a difference.

cnh1995 also makes an interesting point, that you can't wind a coil with several turns and have them all in exactly the same place. The best you might manage is a series of concentric short solenoids each with a slightly larger diameter. I can't agree that the standard formula is exactly correct in either situation, but in practical terms it is probably pretty close in most cases.

OK, I think I understood your point of view, but my answer as a whole is correct or not?

 

[Merlin3189]

Well a) is fine.
I haven't answered b). Everything down to the last 2 lines looks ok and the formula looks familiar, but I don't really remember this quantitative magnetic stuff very well!

With Tesla, Oersted, Gauss, Maxwell, Gilbert and heaven knows how many other named units, not to mention 4pi appearing and disappearing apparently at random, I gave up on it. I try to follow calculations with Amps, metres, Volts, Newtons and so on, but when I see weird units, I stop.
I have the same problem with light - Trollands (both scotopic and photopic), Lamberts, lux, lumens, candels, candellas, candellabras, - and just about any American unit like foot pounds per gallon!

 

[cnh1995]

my answer as a whole is correct or not?

Yes.