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An iron bar of length 10 cm and diameter2cm is placed in a magnetic field of intensity 1000Am^(-1) with its length parallel to the direction of the field. Determine the magnetic moment produced in the bar if permeability of its material is 6.3 xx 10^(-4) TmA^(-1). |
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Answer» SOLUTION :Here `H=1000 Am^(-1) , MU = 6.3 xx 10^(-4) TmA^(-1)` , `l=10 cm , d=2 cm ` Radius of the iron bar = 1CM = `10^(-2)m` we known that , `mu= mu_0 ( 1 + chi_m)` `implies chi_m = (mu)/(mu_0) -1 = (6.3 xx 10^(-4))/(4pi xx 10^(-7))-1` `=500.6` Intensity of magnetisation , ` I = CHIH = 500 .6 xx 1000` `=5 xx 10^(5) Am^(-1)` Magnetic moment , ` M=IV = 5 xx 10^(5) xx pi xx (10^(-2))^(2) xx 0.1 = 5 pi Am^(2)` |
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