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To calculate the inductance of a home made inductor simply take the number of turns,the magnetic flux linkage and the current and use the inductance formula.
Turns-Ratio (TR) = Npri/Nsec Inductance-Ratio (LR) = (Npri/Nsec)^2 = (TR)^2 Turns ratio (TR) = (Npri/Nsec) Inductance ratio (LR) = (Lpri/Lsec) = (Npri/Nsec)^2 = TR^2 TR = SQRT[ Lpri/Lsec ]
An inductor has two properties. The first is resistance(measured in ohms), which is due to the length, cross-sectional area, and resistivity of the conductor from which it is wound. The second is inductance (measured in henrys), which is due to the length of the inductor, its cross-sectional area, the number of turns, and the permeability of its core.The inductor's resistance limits the value of current flowing through the inductor. The inductor's inductance opposes any change in current.
start winding is thicker with less number of winding whereas run winding is less thicker than start winding but with more number of winding. As we know inductance depends upon numbers of turns of winding so run winding will have more inductance.
copper wire diameter,length and numbers of turns for 20 micro Henry inductance and 1600 Amp current
"The magnetic field produced by each turn interacts with the field of other turns and multiplies the effect, causing the inductance of a coil of wire to increase by the number of turns (N) squared. Therefore, if you double the number or turns, you quadruple the inductance."
In addition to the number if turns, the inductance also depends on the length and diameter of the winding, the pitch (number of turns per inch), and the material of the core if there is one. Search on line and find an empirical formula for the inductance of a finite coil, and then work to tweak the other parameters to alter the inductance as required.
radius of coil....number of turns
To calculate the inductance of a home made inductor simply take the number of turns,the magnetic flux linkage and the current and use the inductance formula.
the product of number of turns and flux through the coil ........by maherbano
Yes, even a straight piece of wire has inductance. One metre (or yard) of household flex has about 800nH (0.8uH) of inductance. Winding wire into coils makes them more compact, and multiplies the inductance of the assembly through coupling between turns. Using straight pieces of wire, the inductance is proportional to the length, but in a coiled inductor with 100% coupling, the inductance is proportional to the turns (length) squared! In many applications, iron or ferrite cores are used to further increase the inductance.
You need to shape it into a series of turns, as you have probably seen diagrams of a solenoid. Make as many turns as you can, but don't make the turns too tight. The inductance will be proportional to the number of turns squared, and to the area of each turn, and inversely to the length. If you put a metal rod with high permeability inside this will increase the inductance, but don't let it stick out of the ends.
... increase.
Turns-Ratio (TR) = Npri/Nsec Inductance-Ratio (LR) = (Npri/Nsec)^2 = (TR)^2 Turns ratio (TR) = (Npri/Nsec) Inductance ratio (LR) = (Lpri/Lsec) = (Npri/Nsec)^2 = TR^2 TR = SQRT[ Lpri/Lsec ]
A coil doesn't produce electrical energy, but it can store it. For a given current (amperes), the energy storage in a coil is proportional to the coil's inductance, which in turn depends on the coil's length, diameter, and number of turns. With everything else staying constant, the coil's energy storage capacity increases when the number of turns increases.
As the number of turns in the coil increases, the strength of the electromagnet increases.
There are four basic factors of inductor construction determining the amount of inductance created. These factors all dictate inductance by affecting how much magnetic field flux will develop for a given amount of magnetic field force (current through the inductor's wire coil): 1. Number of turns in the coil (N) 2. Length of coil (l) 3. Cross sectional area of coil (A) 4. Material (nature, or permeability) of coil: u(Greek meu) Inductance, L=(N^2 . u.A)/l