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

Question

"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
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A = 1.96 B = 10.05

Answer 1

By definition, the flux $\Phi_B$ of magnetic induction $B$ through the surface $S$ is (if the magnetic induction is constant):

$$\Phi_B = B_n \cdot S,$$

where $B_n$ 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:

$$B = B_n = \Phi_B/S = \dfrac{10.05\cdot 10^{-4}Wb}{1.96\cdot 10^{-4}m^2} = 5.13 T$$

The intensity of magnetising field is:

$$H = \dfrac{B}{\mu_r\mu_0},$$

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

$$H = \dfrac{5.13T}{1000\cdot 1.26\cdot10^{-6}H/m} =4.07\cdot 10^3 \dfrac{A}{m} ,$$

By definition, the magnetisation then will be:

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

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