Question

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?

Answer 1

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

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

$E_{max}=5.565 V$