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,[tex]\omega=34rad/s[/tex]
We have to find the peak emf of the generator.
[tex]N=\frac{l}{2\pi r}=\frac{10}{2\pi\times 0.14}=11[/tex]
[tex]A=\pi r^2=\pi (0.14)^2=0.062m^2[/tex]
Peak(maximum) induced emf of generator=[tex]E_{max}=NBA\omega[/tex]
Using the formula
[tex]E_{max}=11\times 0.24\times 0.062\times 34[/tex]
[tex]E_{max}=5.565 V[/tex]
