How Do You Choose Wire Type and Calculate Potential Difference for a Solenoid?

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Homework Help Overview

The discussion revolves around constructing a solenoid with specific parameters, including a length of 20 cm and a target magnetic field strength of approximately 1.5 MN/C. Participants are considering two types of wire with different diameters and current-carrying capacities to determine which is more suitable for the solenoid's construction and how to calculate the required potential difference across the coil.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the number of turns of wire, current, and magnetic field strength. There are attempts to calculate the number of turns possible with each wire type and to understand how these relate to the solenoid's requirements. Questions arise regarding the calculations and the underlying equations, particularly concerning the concept of turn density.

Discussion Status

There is active engagement with various calculations and interpretations of the problem. Some participants have provided insights into the equations involved, while others express confusion about specific concepts, such as turn density and its implications for the current requirements. The discussion is ongoing, with multiple lines of reasoning being explored without a clear consensus on the next steps.

Contextual Notes

Participants note the need to consider the current limits of each wire type and how that affects the magnetic field strength. There is also mention of the potential difference that needs to be calculated, which remains unaddressed in the discussion.

jromega3
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Homework Statement


You want to construct a solenoid 20cm long that has an interior magnetic field Strength B (Bc) of about 1.5MN/C. The coil has to be wound as a single layer of wire around a form whose diameter is 3.0 cm. You have two spools of wire handy. #18 wire has a diameter of about 1.02mm and can carry a current of 6.0A before overheating. #26 wire has a diameter of 0.41mm and can carry up to 1.0A. Which kind of wire should you use and why? What potential difference do you want to put across the coil's end?



Homework Equations



B=NI/(cLE0)...or simplified as about 377NI/L.

The Attempt at a Solution



377NI/L must equal about 1.5X10^6N/C...so NI/L must equal 3979. I Have an I...and don't really know where to go from here...and not even sure if this is the right path to take. Not really understanding the question? So I unwind the wires and see how many of them I'd need to fill the 3cm diameter? Any help would be appreciated.
 
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Your L is constant isn't it, so I think you are primarily concerned with the N*I product.

So how many turns can you get along 20 cm using each wire?
Then which of those has the rating to satisfy the current requirement for your target B?
 
ah thanks. Ok so I got N of #18 to be 133, and N of #26 to be 797.
So that's the number of turns. I'm now stuck with how to advance from here.
I assume it has something to do with diameters and am probably missing the obvious here but not seeing how I can relate # of turns to satisfying the requirements of what I'm assuming are diameters.
Thanks again.
 
377NI/L must equal about 1.5X10^6N/C...so NI/L must equal 3979.

So where did that come from?
 
jromega3 said:
ah thanks. Ok so I got N of #18 to be 133, and N of #26 to be 797.
So that's the number of turns. I'm now stuck with how to advance from here.
I assume it has something to do with diameters and am probably missing the obvious here but not seeing how I can relate # of turns to satisfying the requirements of what I'm assuming are diameters.
Thanks again.

Why isn't the number of turns of #18 = 20 cm/1.02mm ?
Similarly for the #26.
 
Those numbers came from B=NI/(cLE0).
Anyway, using those number of turns I'd get 196 and 488 for 18 and 26, respectively.
I guess I just don't see WHAT I'm supposed to be solving for here?
Using those values for N, I get a B of 1.1MN/C for 18 and 0.18MN/C for 26. Neither of which are, or even realistically close to, the 1.5MN/C.
 
jromega3 said:
Those numbers came from B=NI/(cLE0).

I guess I'm not familiar with what that equation is.
 
It's an equation given in my book for the magnetic field inside a solenoid, where B is some special magnetic Field the author uses to make units N/C...and is basically the regular magnetic field times the speed of light.
E0 is epsilon zero too if that wasn't clear.
 
OK. I get what you're doing now.

You've calculated the B field correctly from the units given.

So set 3981 = N/L*I = n*I where n is your turn density.

The turn densities of the 2 wires are 1/D

For the 1.02 mm that comes to 980.4 turns/m and the .41 mm gives 2439 turns/m

One equation then is I = 3981/980.4 < 6a

The other is I = 3981/2439 > 1a
 
  • #10
LowlyPion said:
OK. I get what you're doing now.

You've calculated the B field correctly from the units given.

So set 3981 = N/L*I = n*I where n is your turn density.

The turn densities of the 2 wires are 1/D

For the 1.02 mm that comes to 980.4 turns/m and the .41 mm gives 2439 turns/m

One equation then is I = 3981/980.4 < 6a

The other is I = 3981/2439 > 1a
Lovely. Wow was I stupid. Guess I got so flustered I didn't even understand I was looking for a current below the heating level instead of just solving for B with the info I had like I was doing.
Thanks a lot. Truly appreciated.
 
  • #11
LowlyPion said:
So set 3981 = N/L*I = n*I where n is your turn density.

The turn densities of the 2 wires are 1/D

For the 1.02 mm that comes to 980.4 turns/m and the .41 mm gives 2439 turns/m

One equation then is I = 3981/980.4 < 6a

The other is I = 3981/2439 > 1a

I was following this problem until the concept of turn density was introduced. How does the turn density N/L equal 1/D, assuming that D stands for density of the wire?
 
  • #12
Hey.

I noticed that the complete question was not answered. Anyone know how to answer what the potential difference across the wire needs to be?

Thanks.
 

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