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Andrew [12]
3 years ago
9

The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 10.0 m

. The magnetic field of the generator is 0.24 T, and the coil rotates at an angular speed of 34 rad/s. What is the peak emf of this generator?
Physics
1 answer:
adell [148]3 years ago
3 0

Answer:

5.565 V

Explanation:

Radius of coil of generator=r=0.14 m

Length of wire=l=10 m

Magnetic field,B=0.24 T

Angular speed,\omega=34rad/s

We have to find the peak emf of the generator.

N=\frac{l}{2\pi r}=\frac{10}{2\pi\times 0.14}=11

A=\pi r^2=\pi (0.14)^2=0.062m^2

Peak(maximum) induced emf of generator=E_{max}=NBA\omega

Using the formula

E_{max}=11\times 0.24\times 0.062\times 34

E_{max}=5.565 V

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