Magnetic field inside a solenoid

[geolohs]

Could someone check my calculations?

Homework Statement

Calculate intensity of magnetic field inside of solenoid, who is made from 1mm thick copper wire with 28 twines(n=28). Solenoid is 25 cm long, diameter - 4 cm, and it`s connected to 24V DC voltage.
U=24V ; n=28 ; ρ for copper =0,0175 (ohm*mm2)/m

Homework Equations

H= i n / L
I=U/R
ρ= R*S/L
R=ρ*L/S

The Attempt at a Solution

Perimeter of solenoid=2πr=2*3,14*0,02m=0,126m
Length of solenoid=28 * 0,126 = 3,519 m
S for wire = πr² = 3,14 * 0,25*10^-6 = 0,785*10^-6
Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10^-6=0,078*10⁷
I=U/R=24/0,078*10⁷ = 307,69*10^-6
H= In/L = 307,69*10^-6*28/0,25 = 0,034 A/m

 

[Hesch]

Perimeter of solenoid=2πr=2*3,14*0,02m=0,126m
Length of solenoid=28 * 0,126 = 3,519 m
S for wire = πr² = 3,14 * 0,25*10^-6 = 0,785*10^-6
Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10^-6=0,078*10⁷
I=U/R=24/0,078*10⁷ = 307,69*10^-6
H= In/L = 307,69*10^-6*28/0,25 = 0,034 A/m

Repeated:

Perimeter of solenoid=2πr=2*3,14*0,02m=0,126m
Length of solenoid=28 * 0,126m = 3,519 m
S for wire = πr² = 3,14 * 0,25*10^-6 = 0,785*10^-6
Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10^-6=0,078*10⁷
I=U/R=24/0,078*10⁷ = 307,69*10^-6
H= In/L = 307,69*10^-6*28/0,25 = 0,034 A/m

This is right and this is wrong, and I think it's because you are skipping the units. Then you don't know what you are doing.

The resistance of a 3.519m long copper-wire with a diameter = 1mm will never be 780 kΩ.

 

[geolohs]

I tried it with all units but still don`t get the problem. Why calculations for wire area are wrong? I had to stick to mm², instead of m² or what?

 

[Hesch]

Cross section area of wire = 785*10^-9m².

In the next line you have:

Resistance R=ρ*L/S = 0,0175*3,519 / 0,785*10^-6=0,078*10⁷

Try to write this line with units included, and reduce the unit of the result.

 

[geolohs]

aargh, somehow at the and I get 34285,664 A*mm2/m and this sounds totally wrong :/ How should the units for final answer look like/how big numbers will be?

 

[Hesch]

0,0175 (ohm*mm²)/m
L = 3,519 m
S =0,785*10^-6 m² = 0.785 mm²
R=ρ*L/S = 0.0175 [Ωmm²/m] * 3.519 [m] / 0.785 [mm₂] = 0.0784 Ω

 

[geolohs]

Thanks!
So I=24V/0,078 ohm =306,122 A
Final question - L in H=In/L stands for 3,519m, not 25 cm, right?
So the answer is H=306,122 A * 28 /3,519m = 2435,753 A/m

 

[Hesch]

Final question - L in H=In/L stands for 3,519m, not 25 cm, right?

No. (not quite)

The formula: H = I * n / L is derived from Amperes law:

∫_circulationH⋅ds = N * I

( n / L ) expresses number of turns per length = 28 turns / 0,25 m. The calculated value for H is only valid at the center of the solenoid, so the formula could more exact be written:

H_center = I * ΔN / ΔL

 

[geolohs]

but if I take L as 0,25 m, final answer is 34285,664 A/m, sounds huge

 

[Hesch]

Yes, but N*I = 28*306A = 8568A spread out within 25cm is also a huge current-density. ( turn off the current after 0.1 s, or the copper will melt ).

And, well, it will only be about 0.043T (magnetic induction).

 

[geolohs]

Thank you for big help!