A thin ring with resistance R=10 Ω and self inductance L=0.2 H rotates with constant angular speed ω=100 rad/s in an external uniform magnetic field perpendicular to the rotation axis. As a result, the magnetic flux created by the external field varies with time as

Φ(t)=Φ0cos(ωt)whereΦ0=0.1 Wb.
Because of the resistance of the ring, energy is being continuously dissipated in the system. What average power Pˉ in Watts must develop the external forces to keep the ring rotating at constant angular speed ω (on average).

To find the average power P̅ required to keep the ring rotating at a constant angular speed, we need to consider the energy being dissipated in the system due to the resistance of the ring.

The power dissipated in a resistor is given by the formula P = I^2 * R, where P is the power, I is the current flowing through the resistor, and R is the resistance.

In this case, the current flowing through the ring can be determined using Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a circuit is equal to the negative rate of change of magnetic flux through the circuit.

Mathematically, the induced emf (ε) is given by the formula ε = -dΦ/dt, where Φ is the magnetic flux through the ring.

In this problem, the magnetic flux through the ring is given by Φ(t) = Φ0 * cos(ωt), where Φ0 = 0.1 Wb is the maximum magnetic flux.

Differentiating Φ(t) with respect to time, we get dΦ/dt = -Φ0 * ω * sin(ωt).

Therefore, the induced emf in the ring is ε = -(-Φ0 * ω * sin(ωt)) = Φ0 * ω * sin(ωt).

Using Ohm's law, we can relate the induced emf to the current flowing through the ring. Ohm's law states that the current (I) in a circuit is equal to the emf (ε) divided by the resistance (R).

Therefore, I = ε / R = (Φ0 * ω * sin(ωt)) / R.

To calculate the average power dissipated in the ring, we need to square the current and multiply it by the resistance:

P = (I^2) * R = [(Φ0 * ω * sin(ωt)) / R]^2 * R.

Simplifying the expression, we get:

P = Φ0^2 * ω^2 * sin^2(ωt).

To find the average power, we need to calculate the time average of this expression over one complete period (T) of the oscillation, where T = 2π/ω.

The time average of a sinusoidal function squared over one complete period is 0.5, so we have:

P̅ = 0.5 * Φ0^2 * ω^2.

Substituting the given values, we have:

P̅ = 0.5 * (0.1 Wb)^2 * (100 rad/s)^2 = 0.5 * 0.01 W * 10,000 W = 50 W.

Therefore, the average power P̅ that must develop the external forces to keep the ring rotating at a constant angular speed ω is 50 Watts.