# Magnetic field inside a solenoid

• geolohs
In summary, someone attempted to do a calculation for a solenoid, but they got the wrong answer because they skipped over the units. They also had trouble with the cross-sectional area of the wire. The solution is to use Amperes law and to calculate the Hcenter. This value is only valid at the center of the solenoid, which is why the final answer is 34285,664 A/m.
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 its connected to 24V DC voltage.
U=24V ; n=28 ; ρ for copper =0,0175 (ohm*mm2)/m

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

geolohs said:
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.

Last edited:
geolohs
I tried it with all units but still dont get the problem. Why calculations for wire area are wrong? I had to stick to mm2, instead of m2 or what?

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.

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?

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

geolohs said:
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:

Hcenter = I * ΔN / ΔL

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

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).

Last edited:
geolohs
Thank you for big help!

## 1. What is a solenoid?

A solenoid is a coil of wire that is often used to generate a magnetic field. It is typically cylindrical in shape and contains multiple loops of wire.

## 2. How is a magnetic field created inside a solenoid?

A magnetic field is created inside a solenoid when an electric current passes through the wire coils. The direction of the magnetic field is determined by the direction of the current flow.

## 3. What factors affect the strength of the magnetic field inside a solenoid?

The strength of the magnetic field inside a solenoid is affected by the number of wire coils, the amount of current passing through the coils, and the material of the core inside the solenoid.

## 4. How does the direction of the magnetic field inside a solenoid change with the direction of current flow?

The direction of the magnetic field inside a solenoid is determined by the right-hand rule. If the current flows in the direction of your fingers wrapped around the solenoid, the magnetic field will point in the direction of your thumb pointing out of the solenoid.

## 5. What is the purpose of a solenoid in practical applications?

Solenoids are used in a variety of applications, including electromagnets, speakers, and electronic locks. They can also be used to control the flow of fluids or gases in industrial processes.

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