A 0.35 m2 coil with 50 turns produces a maximum emf of 45 volts in a magnetic field of 0.4 Tesla. What is the speed of rotation of the coil?
Can you show the steps too so I can understand? Thanks
ε=dÔ/dt = d(NBAcosωt)/dt =
= - NBAω sinωt,
ε(max) = NBAω =>
ω=ε(max)/ NBA =45/50•0.4•0.35 =6.43 rad/s
To find the speed of rotation of the coil, we can use Faraday's law of electromagnetic induction. This law states that the induced electromotive force (emf) in a circuit is equal to the rate of change of magnetic flux through the circuit.
The magnetic flux is given by the equation:
Φ = B * A * cos(θ)
where Φ is the magnetic flux, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal vector to the coil.
In this case, we are given the maximum emf (E), the magnetic field (B), and the area of the coil (A). The angle θ is not given, but we can assume it to be 0 degrees (cos(0) = 1), since the coil's plane is perpendicular to the magnetic field.
So, the equation becomes:
E = dΦ / dt
Now, let's find the change in magnetic flux over time.
Since the flux is defined as Φ = B * A * cos(θ), we can write:
Φ = B * A * cos(0) (assuming θ = 0)
Φ = B * A
Differentiating both sides of the equation with respect to time, we get:
dΦ / dt = d(B * A) / dt
The area (A) of the coil does not change, so the derivative of A with respect to time is 0. Therefore, the equation simplifies to:
dΦ / dt = B * dA / dt
However, dA / dt represents the rate of change of the area, which is equal to the rate of change of the angle between the normal vector of the coil and the magnetic field direction (θ). Let's call this angular speed ω (omega).
So, dΦ / dt = B * dA / dt can be further simplified as:
dΦ / dt = B * ω
Now, substituting this equation back into Faraday's law:
E = dΦ / dt
E = B * ω
Finally, we can solve for the angular speed (ω):
ω = E / B
Given that E = 45 volts and B = 0.4 Tesla, we can evaluate ω:
ω = 45 / 0.4
Therefore, the speed of rotation of the coil is 112.5 radians per second.