In a uniform magneitc field of induced B a wire in the form of a semicircle of radius r rotates about the diameter of hte circle with an angular frequency `omega`. The axis of rotation is perpendicular to hte field. If the total resistance of hte circuit is R, the mean power generated per period of rotation is

Options

A. `(Bpir^(2)omega)/(2R)`
B. `((Bpir^(2)omega)^(2))/(8R)`
C. `((Bpiromega)^(2))/(2R)`
D. `((Bpiromega^(2))^(2))/(8R)`

Explanation

To solve the problem step by step, we will follow the concepts of electromagnetic induction and the formulas related to induced EMF and power.

### Step 1: Understand the Setup
We have a semicircular wire of radius \( r \) rotating in a uniform magnetic field \( B \) with an angular frequency \( \omega \). The axis of rotation is perpendicular to the magnetic field.

### Step 2: Calculate the Area of the Semicircle
The area \( A \) of a semicircle is given by:
\[ A = \frac{1}{2} \pi r^2 \]

### Step 3: Determine the Magnetic Flux
The magnetic flux \( \Phi \) through the semicircular wire is given by:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where \( \theta = \omega t \) (the angle between the magnetic field and the normal to the area). Substituting the area:
\[ \Phi = B \cdot \left(\frac{1}{2} \pi r^2\right) \cdot \cos(\omega t) = \frac{1}{2} B \pi r^2 \cos(\omega t) \]

### Step 4: Calculate the Induced EMF
The induced EMF \( E \) can be calculated using Faraday's law of electromagnetic induction:
\[ E = -\frac{d\Phi}{dt} \]
Differentiating the flux:
\[ E = -\frac{d}{dt} \left(\frac{1}{2} B \pi r^2 \cos(\omega t)\right) \]
Using the chain rule:
\[ E = -\frac{1}{2} B \pi r^2 \cdot (-\omega \sin(\omega t)) = \frac{1}{2} B \pi r^2 \omega \sin(\omega t) \]

### Step 5: Calculate the Instantaneous Power
The instantaneous power \( P \) is given by:
\[ P = \frac{E^2}{R} \]
Substituting the expression for \( E \):
\[ P = \frac{\left(\frac{1}{2} B \pi r^2 \omega \sin(\omega t)\right)^2}{R} = \frac{B^2 \pi^2 r^4 \omega^2 \sin^2(\omega t)}{4R} \]

### Step 6: Calculate the Mean Power
To find the mean power over one complete cycle, we need to take the average of \( \sin^2(\omega t) \). The average value of \( \sin^2(\omega t) \) over one period is \( \frac{1}{2} \):
\[ P_{\text{mean}} = \frac{B^2 \pi^2 r^4 \omega^2}{4R} \cdot \frac{1}{2} = \frac{B^2 \pi^2 r^4 \omega^2}{8R} \]

### Final Answer
Thus, the mean power generated per period of rotation is:
\[ P_{\text{mean}} = \frac{B^2 \pi^2 r^4 \omega^2}{8R} \]