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3 years ago

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An AC generator consists of 6 turns of wire. Each turn has an area of 0.040 m2. The loop rotates in a uniform field (B = 0.20 T)

at a constant frequency of 50 Hz. What is the maximum induced emf?

1 answer:

Alja [10]3 years ago

5 0

Answer:

The maximum induced emf is 15.08 V

Explanation:

Given;

number of turns of the generator, N = 6 turns

area of the loop, A = 0.04 m

magnetic field of the loop, B = 0.2 T

frequency of loop, f = 50 Hz

The maximum induced emf is given by;

E = NBAω

Where;

ω is the angular speed = 2πf

E = NBA(2πf)

E = 6 x 0.2 x 0.04 x (2 x 3.142 x 50)

E = 15.08 V

Therefore, the maximum induced emf is 15.08 V

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Explanation:

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$F_N=80\times 9.8\times cos25^{\circ}$

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$W_l=331.33\,N$

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$F=f+W_l$

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$F=437.91\,N$

<em>Hence the work done:</em>

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$W=96340.80\,J$

<em>Now power:</em>

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$P=698.12\,W$

<u>So, power required for 30 such bodies:</u>

$P=30\times 698.12$

$P=20943.65\,W$

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