You want to use a windmill to generate electricity for your home. You decide to use Earths magnetic field which has a horizontal component B=10^-4 T to an induce emf. You design a conduction loop with 200 turns and surface area of 0.1m^2. The rod is 0.5m wide. On the day you you test your windmill the wind speed was measured to be 100km/hr, assume the windmill rotates at this spped. What is the maximum voltage induced by the wind
100 km/hr is not a rotational speed. Where is the rod located? How is the loop oriented with respect to the field?
I don't get the picture here.
To calculate the maximum voltage induced by the wind in the windmill, we need to use Faraday's law of electromagnetic induction.
Faraday's law states that the induced electromotive force (emf) in a closed loop is equal to the rate of change of magnetic flux passing through that loop.
First, let's calculate the magnetic flux passing through the loop. The magnetic flux (Φ) through a loop is given by the equation Φ = B * A * cos(θ), where B is the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.
In this case, the horizontal component of Earth's magnetic field (B) is given as 10^-4 T, the surface area of the loop (A) is 0.1 m^2, and the angle between the magnetic field and the normal to the loop (θ) is 0 degrees.
Φ = B * A * cos(θ)
Φ = (10^-4 T) * (0.1 m^2) * cos(0°)
Φ = 10^-5 Wb
Next, we need to calculate the rate of change of magnetic flux. Since the windmill rotates at a speed of 100 km/hr, we need to convert this speed into meters per second.
100 km/hr = (100,000 m) / (3600 s)
100 km/hr ≈ 27.78 m/s
The windmill completes one revolution (2π radians) in one second, so the angular velocity (ω) of the windmill is 2π rad/s.
Now, the rate of change of magnetic flux (dΦ/dt) is equal to B * A * ω * sin(θ), where ω is the angular velocity and θ is the angle between the magnetic field and the normal to the loop.
dΦ/dt = B * A * ω * sin(θ)
dΦ/dt = (10^-4 T) * (0.1 m^2) * (2π rad/s) * sin(0°)
dΦ/dt = 6.28 × 10^-4 Wb/s
Using Faraday's law, the induced emf (V) in the loop is equal to the rate of change of magnetic flux.
V = dΦ/dt
V = 6.28 × 10^-4 V
Finally, since you have a conduction loop with 200 turns, the maximum voltage induced in the windmill is equal to the product of the induced emf and the number of turns.
Maximum voltage induced = V * Number of turns
Maximum voltage induced = (6.28 × 10^-4 V) * 200
Maximum voltage induced ≈ 0.126 V
Therefore, the maximum voltage induced by the wind in the windmill is approximately 0.126 volts.