Step 1: Understanding the Concept:
A rotating loop in a magnetic field experiences a change in magnetic flux, which induces an EMF according to Faraday's Law. This induced EMF produces a current, leading to power dissipation in the resistance of the wire.
Step 2: Key Formula or Approach:
1. Magnetic Flux: \( \phi = \vec{B} \cdot \vec{A} = BA \cos \theta \).
2. Induced EMF: \( e = -\frac{d\phi}{dt} \).
3. Instantaneous Power: \( P = \frac{e^2}{R} \).
4. Average (Mean) Power: \( P_{avg} = \frac{1}{T} \int_{0}^{T} P dt \).
Step 3: Detailed Explanation:
The area of the semi-circular loop is \( A = \frac{\pi r^2}{2} \).
The loop rotates with angular velocity \( \omega \), so the angle is \( \theta = \omega t \).
Flux \( \phi = B \left( \frac{\pi r^2}{2} \right) \cos(\omega t) \).
Induced EMF \( e = -\frac{d\phi}{dt} = B \left( \frac{\pi r^2}{2} \right) \omega \sin(\omega t) \).
Instantaneous Power \( P_{ins} = \frac{e^2}{R} = \frac{(B \pi r^2 \omega / 2)^2 \sin^2(\omega t)}{R} = \frac{(B \pi r^2 \omega)^2}{4R} \sin^2(\omega t) \).
To find the mean power over a period, we use the fact that the average value of \( \sin^2(\omega t) \) over one complete cycle is \( \frac{1}{2} \).
\[ P_{avg} = \frac{(B \pi r^2 \omega)^2}{4R} \times \text{Average}(\sin^2 \omega t) = \frac{(B \pi r^2 \omega)^2}{4R} \times \frac{1}{2} \]
\[ P_{avg} = \frac{(B \pi r^2 \omega)^2}{8R} \]
Comparing this with the given form \( \frac{(B\pi r^2 \omega)^2}{xR} \), we find \( x = 8 \).
Step 4: Final Answer:
The value of x is 8.