A solenoid is wound with a single layer of insulated copper wire of diameter 2.800 mm and has a diameter 5.000 cm and is 1.500 m long. Assume that the adjacent wires touch and that the insulation thickness is negligible. what is the inductance per meter(H/m) for the solenoid near its center.
Isn't there a standard formula for this? All you have to really calculate is the turns. I think that will be close to 1.5/.0028
check my thinking.
This is a two part question the first part was to find the number of turns which is 535 like you said.
L=(mewsub0*N^2*A)/length
A= .5 pi r^2
N=number of turns=535
mewsubO= 4 pi 10^-7
thats the formula to use but I'm not getting a correct answer any suggestions?
How can area be 1/2 PI r^2? That is not what I memorzed in the sixth grade.
haha thanks bobpursley you are right that's the problem
To calculate the inductance per meter (H/m) of a solenoid, you can use the formula:
L = (μ₀ * N² * A) / l
Where:
- L is the inductance in henries (H)
- μ₀ is the permeability of free space (4π x 10⁻⁷ H/m)
- N is the number of turns
- A is the cross-sectional area of the solenoid
- l is the length of the solenoid
First, we need to calculate the number of turns in the solenoid. To do that, we can divide the length of the solenoid by the thickness of the wire (diameter):
Number of Turns (N) = Length (l) / Wire Diameter
N = 1.500 m / 2.800 mm
Next, we need to calculate the cross-sectional area (A) of the solenoid. We can use the formula for the area of a circle:
Area (A) = π * (Radius)²
First, let's convert the diameter of the solenoid from centimeters to meters:
Solenoid diameter = 5.000 cm = 0.05000 m
Radius = Solenoid diameter / 2
Next, we can calculate the area:
Area (A) = π * (0.05000 m / 2)²
Finally, we can substitute the values into the formula for inductance:
L = (4π x 10⁻⁷ H/m) * (N²) * (A) / l
Now you can perform the calculations to find the inductance per meter of the solenoid near its center.