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Derive an Expression for instantaneous induced emf in an A.C generator |
Answer» Solution : In the above figure N and S are poles of a magnet, `theta` is the angle between the direction of magnetic FIELD B and area vector A. When the COIL is rotated in the magnetic field, the flux linked with the coil varies. At any instant of time .t., `A cos theta` is the component of area vector along the direction B. The magnetic flux linked iwth the coil at any instant of time .t. is given by `phi_(B)= B XX` component of area vector along the field direction. For 1 turn `phi_(B)= BA cos theta` For n turns `phi_(B)= nAB cos theta` `phi_(B)= nAB cos omega t` ....(1) `[because theta= omega t]` where `.omega.` is the angular velocity of the coil at time t From the Faraday.s second law, `e= - (d phi)/(dt)` `e= - (d)/(dt)[ n AB cos omega t]` [From (1) `phi= nAB cos omega t`] `e= (-nAB)[-sin omega t] xx omega` `e= nA B omega sin omega t` `e= e_(0) sin (omega t)` where `e_(0)` is the PEAK value of emf `=nAB omega` |
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