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A coil of 325 turns, with a coil side of active length 510mm, rotates in a uniform magnetic field between the poles of an electromagnet which give a field of flux density 0.033T. If the conductor speed is 15.2 m/s, determine the maximum e.m.f generated in the coil and the e.m.f generated at the instant when the conductors are 400 after the position for maximum e.m.f.

A coil of 325 turns, with a coil side of active length 510mm, rotates in a uniform-example-1
User Soma
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1 Answer

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Given that the number of turns of the coil is n = 325

The length is l = 510 mm = 0.51 m

The flux density is B = 0.033 T

The speed of the conductor is v = 15.2 m/s

The maximum induced emf will be


\begin{gathered} e_(\max )=nBlv\sin 90^(\circ) \\ =325*0.033*0.51*15.2*1 \\ =83.1402\text{ V} \end{gathered}

The emf when the angle is 40 degrees will be


\begin{gathered} e=nBlvsin40^(\circ) \\ =325*0.033*0.51*15.2*0.643 \\ =\text{ 53.46 V} \end{gathered}

User SKManX
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