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Pregunta: An alternating current generator has 12 turns of wire, a radius of 12 cm, and a wire resistance of 15 Q2. The coil rotates in a 300 mT magnetic field at an angular frequency of 314 rad/s. Calculate the maximum induced emf in the coil. or to a.5.67 A b. 3.4 A c. 6.70 Ad. 21.3A
An alternating current generator has 12 turns of wire, a radius of 12 cm, and a wire resistance of 15 Q2. The coil rotates in a 300 mT magnetic field at an angular frequency of 314 rad/s. Calculate the maximum induced emf in the coil. or to
a.5.67 A
b. 3.4 A
c. 6.70 A
d. 21.3A

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Solución
Paso 1

As we know that, Maximum induced emf = NABw Here,N= number of turns = 12 A= area = π* 0.12 *0.12 = 0.045 m^2 B= magnitude of magnetic field= 300 mT w= 314 rad/S After putting all values in above formula

$\begin{matrix} \begin{aligned} E m & = 12 \times π \times 0.12 \times 0.12 \times 300 \times 10^{- 3} \times 314 \\ E m & = 51.14 V \end{aligned} \end{matrix}$
So, maximum induced emf is 51.14 V
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de 300 mi a ura trecuencia angular de 314 radis. Calcula la fem liducida maximn en la bobina.

14567 A

(1)

70 A

12131