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adell [148]
3 years ago
5

You need to design a 60.0-Hz ac generator that has a maximum emf of 5500 V. The generator is to contain a 150-turn coil that has

an area per turn of 0.85 m 2 . What should be the magnitude of the magnetic field in which the coil rotates?
Physics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

So magnetic field will be equal to 0.1144 T  

Explanation:

We have given frequency f = 60 Hz

Maximum emf e = 5500 volt

Number of turns N = 150

Area A=0.85m^2

Emf generated in ac generator is given e=NBA\omega sin(\omega t)

For maximum emf sin(\omega t)=1

So maximum emf will be equal to e=NBA\omega

B=\frac{e}{NA\omega }=\frac{5500}{150\times 2\times 3.14\times 60\times 0.85}=0.1144T

So magnetic field will be equal to 0.1144 T

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