Question

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?

Answer 1

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

Answer 2

The magnitude of the magnetic field in which the coil rotates is; B = 0.1144 T

We are given;

Frequency; F = 60 Hz

Maximum Emf; ε_max = 5500 V

Number of turns of coil; N = 150 turns

Area; A = 0.85 m²

Formula for maximum emf is;

ε_max = NBAω

Where;

ω = 2πf

Thus;

ε_max = NBA(2πf)

Making B the magnetic field the subject gives;

B = ε_max/(NA × 2πf)

Plugging in the relevant values;

B = 5500/(150 × 0.85 × 2π × 60)

B = 0.1144 T

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