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authorpriyanka2015-06-24 15:03:17 +0530
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+// chapter 2 example 1
+//------------------------------------------------------------------------------
+clc;
+clear;
+// µr1 = 3; // relative permeability of region 1
+// µr2 = 5; // relative permeability of region 2
+// H1 = (4ax + 3ay -6az)A/m; Magnetic field intensity
+// Therefore B1 = µoµr1H1
+// = µo(12ax + 9ay -18az)A/m
+// since normal component of (B) is continuous across the interface
+// Therefore, B2 = µo[12ax + 9(µr2/µr1)ay -18(µr2/µr1)az]
+// = µo[12ax + 15ay - 30az]
+// H2 = [12/5ax + 15/5ay - 30/5az]A/m
+// H2 = (2.4ax + 3ay - 6az)
+H2 = sqrt(2.4^2 + 3^2 + 6^2);
+
+// output
+mprintf('Magnetic field intensity in region- 2 = %3.2f A/m',H2);
+//------------------------------------------------------------------------------