The coil of a generator has a radius of 0.12 m. When this coil is unwound, the wire from which it is made has a length of 5.8 m. The magnetic field of the generator is 0.21 T, and the coil rotates at an angular speed of 26 rad/s. What is the peak emf of this generator?

Answer in V

Use the information on wire length and radius to get the number of turns, N.

5.8 meters = 2*pi*R*N
N = 7.7 Let's call it 7 turns. The other 0.7 m might be used for connections.

Maximum Flux = Phi = N*B*A
= 7*0.21*pi*R^2
= 0.067 Webers

Peak EMF = w*Phi= 1.7 volts
where w is the angular velocity of the coil.

To find the peak emf (electromotive force) of the generator, we can use Faraday's law of electromagnetic induction, which states that the emf induced in a coil is proportional to the rate of change of magnetic flux through the coil.

The magnetic flux (Φ) through the coil can be calculated using the formula: Φ = B * A * N * cos(θ), where B is the magnetic field, A is the area of the coil, N is the number of turns in the coil, and θ is the angle between the magnetic field and the normal to the coil.

Given:
Radius of the coil (r) = 0.12 m
Length of the wire (L) = 5.8 m
Magnetic field (B) = 0.21 T
Angular speed (ω) = 26 rad/s

To find the area of the coil (A), we need to know the shape of the coil. Assuming it is a flat, circular coil, we can calculate the area using the formula: A = π * r^2.

A = π * (0.12 m)^2
A = 0.0452 m^2

Now, let's find the number of turns in the coil (N). We can do this by dividing the length of the wire (L) by the circumference of the coil.

Circumference of the coil = 2 * π * r
Circumference of the coil = 2 * π * 0.12 m
Circumference of the coil = 0.753 m

N = L / Circumference of the coil
N = 5.8 m / 0.753 m
N = 7.7

Now, we need to find the angle (θ) between the magnetic field and the normal to the coil. Assuming the magnetic field is perpendicular to the coil (θ = 0 degrees), the cosine of 0 degrees is 1.

θ = 0 degrees
cos(θ) = 1

Now, we can calculate the magnetic flux (Φ):

Φ = B * A * N * cos(θ)
Φ = 0.21 T * 0.0452 m^2 * 7.7 * 1
Φ ≈ 0.074 T∙m^2

Finally, we can calculate the peak emf (ε) using the formula: ε = -N * dΦ/dt, where dΦ/dt is the rate of change of the magnetic flux.

Since the coil rotates at an angular speed of 26 rad/s, the time derivative of the magnetic flux is given by dΦ/dt = -B * A * ω * sin(θ), where ω is the angular speed.

dΦ/dt = -0.21 T * 0.0452 m^2 * 26 rad/s * sin(θ)
dΦ/dt = -0.25 T∙m^2/s

ε = -N * dΦ/dt
ε = -7.7 * (-0.25 T∙m^2/s)
ε ≈ 1.93 V

Therefore, the peak emf of this generator is approximately 1.93 V.