InterviewSolution
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InterviewSolution
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A circular coil of radius 0.08m consisting of 100 turns is carrying a current of0.4A. Calculate the magnitude of the magnetic field i) at the center of the coiland ii) at a point 0.2m from the center of the coil on its axis. |
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Answer» Solution : GIVEN : r=0.08m, n = 100 turns, I = 0.4A i) Magnetic FIELD at the centre of the coil `B = (mu_0 nI)/(2r)` `B = (4pi xx 10^(-7) xx 100 xx 0.4)/(2 xx 0.08)` `B = (502.4 xx 10^(-7))/(0.16)` `B = 3.140 xx 10^(-7)` `B = 3.14 xx 10^(-4) T` ii) At a POINT 0.2m from the centre of the coil on its axis `B = (mu_0)/(4pi) (2PI n I r^2)/((x^2+ r^2)^(3//2))` `B = 10^(-7) xx (2 xx 3.14 xx 100 xx 0.4 xx (0.08)^2)/([(0.2)^2 + (0.08)^2]^(3//2))` `B= (1.607 xx 10^(-7))/((0.04 + 0.0064)^(3//2))` `B= (1.607 xx 10^(-7))/(0.3593)` `B = 4.47 xx 10^(-7) T` |
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