Question #186190

An iron ring uniformly wound with 1200 turns of insulated wire and a magnetizing current of 1.25 A produces a flux of 1 mWb in the air gap. The mean circumference of the ring is 50 cm and cross-sectional area of 20 cm². A radial air gap of 0.5 mm is cut in the ring. Neglecting the effect of magnetic leakage and fringing, caluculate (a) the reluctance of the magnetic circuit and (b) the relative permeability of iron


Expert's answer

(a) the reluctance of the magnetic circuit

H=NIl=1200∗1250∗10−2=2880A/mH= \frac{NI}{l}=\frac{1200 *12}{50* 10^{-2}}=2880 A/m

μr=BH=1∗10−3/20∗10−22800=1/576000\mu_r=\frac{B}{H}=\frac{1*10^{-3}/20*10^{-2}}{2800}= 1/576000

S=lμA=0.5×10−31/576000∗4π∗10−7∗20∗10−4=1.1459×1011A/WbS=\frac{l}{\mu A}=\frac{0.5 \times 10^{-3}}{1/576000 *4 \pi *10^{-7}*20*10^{-4}}=1.1459 \times10^{11} A/Wb

(b) the relative permeability of iron

μr=BH=1∗10−3/20∗10−22800=1/576000=0.00000173611\mu_r=\frac{B}{H}=\frac{1*10^{-3}/20*10^{-2}}{2800}= 1/576000=0.00000173611


LATEST TUTORIALS
APPROVED BY CLIENTS