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2 years ago

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An AC generator has 80 rectangular loops on its armature. Each loop is 12 cm long and 8 cm wide. The armature rotates at 1200 rp

m about an axis parallel to the long side. If the loop rotates in a uniform magnetic field of 0.30 T, which is perpendicular tothe axis of rotation, what will be the maximum output voltage of this generator?

1 answer:

Sauron [17]2 years ago

5 0

Answer: 28.96 V

Explanation:

Given

No of loops on the armature, N = 80

Length of the loop, l = 12 cm = 0.12 m

Width of the loop, b = 8 cm = 0.08 m

Speed of the armature, 1200 rpm

Magnetic field of the loop, B = 0.30 T

To solve this, we use the formula

V(max) = NBAω

Where,

A = area of loop

A = l*b = 0.12 * 0.08

A = 0.0096 m²

ω = 1200 rpm = 1200 * 2π/60 rad/s

ω = (1200 * 2 * 3.142) / 60

ω = 7540.8 / 60

ω = 125.68 rad/s

Substituting the values into the formula

V(max) = NBAω

V(max) = 80 * 0.30 * 0.0096 * 125.68

V(max) = 80 * 0.362

V(max) = 28.96 V

Therefore, the maximum output voltage of the generator would be 28.96 V

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