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A coil of a tangent galvanometer of diametre 0.24 m has 100 turns. If the horizontal component of Earth's magnetic field is 25 xx 10^(-6) T then, calculate the current which gives a deflection of 60^(@) . |
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Answer» Solution :The DIAMETER of the COIL is 0.24 m. therefore, radius of the coil is 0.12m. Number of turns is 100 turns. Earth.s magnetic field is 25 `xx 10^(-6)` T DEFLECTION is `THETA = 60^(@) rArr tan 60^(@) = sqrt(3) = 1.732` I = `(2 R B_(H))/(mu_(0)N) tan theta = (2 xx 0.12 xx 25 xx 10^(-16))/(4 xx 10^(-7) xx 3.14 xx 100) xx 1.732 = 0.82 xx 10^(-1) `A I = 0.082 A |
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