Answer:
R = 7.854 x 10⁵ anpert turns / Wb
Relative permeability = 405.3
Explanation:
Detailed explanation is given in the attached document.
Answer: 85398.16, 405.28473.
Explanation:
We are given that the number of turns on the core is 500 2 , the cross sectional area is :
A=5cm^{2}({1meter}/{100cm})^{2} =0.0005meters^{2}
and the length of the core is l=20cm ({1meter}/{100cm})= 0.2meters .
In this solution, we are meant to neglect the resistance of the coil , and the current through the coil is I=1amPrms when the voltage applied across it is:
V=120voltsrms at f=60Hz. From this, We can calculate the inductance(L) whiof the coil (a coil have an inductance value of one Henry when an electromotive force of one volt is induced in the coil were the current flowing through the said coil changes at a rate of one ampere/second).
The natural frequency of the applied voltage is:
ω =2π f=2π(60)=120π{radians}/{second} .
The inductive reactance of the coil is equal to X=ω L=120π L . We then know that current is :
I=V/X
=I =20/120πL
L=120/120π
=1/π henries .
For reluctance (R) (which is a unit measuring the opposition to the flow of magnetic flux within magnetic materials and is analogous to resistance in electrical circuits). Looking at the relationship between inductance and reluctance . You will note that it is :
L=n^2/R .
We can use this relationship to find reluctance for our closed iron core coil :
L=1/π = 500^2/R
R=500^2π =785398.16{amps}/{volt-seconds}}
We can therefore use the other equation for reluctance.
R=1/μ A= 1/μA or μRA
To calculate the relative permeability of the core :
R=500^2π
=0.2/(4π ×10^-7)/ μR(0.0005)
μR=0.2/(4π ×10^-7)/ 500^2π(0.0005)
= 405.28473
