Answer to Question #117321 in Electricity and Magnetism for shreevathsa BK

Question #117321
"An iron rod of length 0.5m has a cross sectional area of A x 10-4 m2. If the flux of magnetic induction through the bar is B x 10-4 Weber and relative permeability of iron is 1000, calculate, Magnetisation M and Intensity of magnetising field H
"
A = 1.96 B = 10.05
1
Expert's answer
2020-05-24T18:01:01-0400

By definition, the flux ΦB\Phi_B of magnetic induction BB through the surface SS is (if the magnetic induction is constant):


ΦB=BnS,\Phi_B = B_n \cdot S,

where BnB_n is the normal to the surface component of the electromagnetic induction. As far as the rod is long and rather thin, one can assume, that the magnetic induction is constant inside the rod and directed along it's axis. Thus, the magnetic induction will be:


B=Bn=ΦB/S=10.05104Wb1.96104m2=5.13TB = B_n = \Phi_B/S = \dfrac{10.05\cdot 10^{-4}Wb}{1.96\cdot 10^{-4}m^2} = 5.13 T

The intensity of magnetising field is:


H=Bμrμ0,H = \dfrac{B}{\mu_r\mu_0},

where μr=1000\mu_r = 1000 is the relative permeability of iron and μ0=1.26106H/m\mu_0 = 1.26\cdot 10^{-6} H/m is the magnetic permeability of free space. Thus:


H=5.13T10001.26106H/m=4.07103Am,H = \dfrac{5.13T}{1000\cdot 1.26\cdot10^{-6}H/m} =4.07\cdot 10^3 \dfrac{A}{m} ,

By definition, the magnetisation then will be:


M=1μ0BH=5.131.261064.071034.07106AmM = \dfrac{1}{\mu_0}B-H = \dfrac{5.13}{1.26\cdot 10^{-6}}-4.07\cdot 10^3 \approx 4.07\cdot 10^6 \dfrac{A}{m}


Answer. M = 4.07*10^6 A/m, H = 4.07*10^3 A/m.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment