Question

A coil is wound uniformly on an iron core. The relative permeability of the iron is 1400 .The length of the magnetic circuit is 70m. The cross - sectional area of the core is 5 m^2 the coil has 500 turns ,calculate the reluctance(s),inductance(l).​

Answer 1

HELLO DEAR,

GIVEN:-

relative permeability  μr = 1400

cross section area A = 5m²

N = 500turns

length of the circuit l is 70m

we know,

Reluctance is defined as the opposition offered to that magnetic circuit to the passage of magnetic flux through it (the ratio of the magnetic potential difference to the corresponding flux).

so,  reluctance (S) = l/μA

    μ = (μo)(μr)

    μ = (4π×10^-7)(1400)

    μ = 56π×10^-5

 

$\bold{\therefore S = \frac{70}{56\pi \times 10^{-5}\times 5}}$

=> S = 10/(8π×10^-5×5)

=> S = 10^6/40π

=> S = 10^6/125.6

=> S = 7961.78

=> S =  7.962×10³ A/Wb

similarly,

Inductance(L) is the tendency of an electrical conductor to oppose a change in the electric current flowing through it.

where, L = μN²A/l

$\bold{L = \frac{56\pi \times 10^{-5} \times (500)^2 \times 5}{70}}$

$\bold{L = \frac{8*5*3.14*10^{-5}*25*10^{4}}{10}}$

$\bold{L = 4*3.14*25*10^{-5}*10^{4}}$

$\bold{L = 100*3.14*10^{-1}}\\\\\bold{L=3.14*10 = 31.4 Henry}\\\\\bold{or \;10\pi Henry}$

THANKS.