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A wheel with 15 metallic spokes, each 60 cm long, is rotated at 360 revolutions per minute in a plane normal to the horizontal component of earth's magnetic field. The angle of dip at that place is 60^(@). If the emf induced between rim of the wheel and the axle 400 mV, calculate the horizontal component of earth's magnetic field at the place. How will the induced emf change if the number of spokes is increased ? |
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Answer» Solution :Here number of spokes in wheel n = 15, length of each spoke l = 60 cm = 0.6 m, angular SPEED of wheel `omega = 360 rpm = rps = (360)/60 rps = 6 xx 2pi rad s^(-1) = 12pi rad s^(-1)` and INDUCED emf `varepsilon = 400 mV = 400 xx 10^(-3) V = 0.4 V` As wheel is revolving in a plane normal to the horizontal component `B_(H)`, induced emf between its AXLE and the rim of wheel `|varepsilon| =1/2B_(H)l^(2)omega implies B_(H) = (2varepsilon)/(l^(2)omega) = (2 xx 0.4)/((0.6)^(2) xx (12pi))=5.9xx10^(-2)T` |
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