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Derive an expression for the induced emf developed when a coil of N turns, and area of cross-section A, is rotated at a constant angular speed omega in a uniform magnetic field B. (b) A wheel with 100 metallic spokes each 0.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth's magnetic field. If the resultant magnetic field at that place is 4 xx 10^(-4) T and the angle of dip at the place is 30^(@), find the emf induced between the axle and the rim of the wheel. |
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Answer» Solution :(a) Consider a coil PQRS of N turns, each of area A, held in a uniform magnetic field `vecB`. Let thecoil be rotated at a steady angular velocity w about its own axis. Let at any instant t, normal to the area (i.e., the area vector `vecA`) subtends an angle `theta = omegat` from direction of magnetic field `vecB`. Then, at that moment, the magnetic flux linked with the coil is `phi_(B) = NvecB.vecA = N B A cos theta = cos omega t` `therefore` Induced emf `varepsilon = -(dphi_(B))/DT = - d/dt (NBA cos omegat` `= - N B A d/dt (cos omegat)` `= N B A omega SIN omegat` (b) Here number of spokes N = 100, length of each spoke L = 0.5 m, angular frequency `omega = 120 "rev"//"min" = (120)/60 rps = (120)/60 xx 2pi"rad s"^(-1) = 4pi "rad s"^(-1),` earth.s RESULTANT magnetic field `B_(E) = 4 xx 10^(-4)T` and angle of dip `delta = 30^(@)` As wheel is being rotated in a plane normal to the horizontal component `(B_(H))` of the earth.s magnetic field, hence induced emf between the axle and the rim of the wheel due to any one spoke. 1` varepsilon = 1/2 B_(H) varepsilonl^(2)=1/2B_(E) cos deltaomegal^(2)= 1/2 xx(4xx10^(-4))xx cos30^(@) xx4pi xx(0.5)^(2) =5.4 xx 10^(-4) V` As all the 100 spokes are joined in parallel, the total induced emf will be same as the emf due to one spoke. `implies` Total induced emf `= 5.4 xx 10^(-4)V or 0.54 mV` 
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