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

Options

A. `((B pi r omega)^(2))/(2R)`
B. `((B pi r^(2) omega)^(2))/(8R)`
C. `(B pi r^(2) omega)/(2R)`
D. `((B pi r omega^(2))^(2))/(8R)`

Explanation

To solve the problem, we need to determine the mean power generated per period of rotation for a wire in the form of a semicircle rotating in a uniform magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a semicircular wire of radius \( r \) rotating about its diameter with an angular frequency \( \omega \) in a uniform magnetic field \( B \). The total resistance of the circuit is \( R \).

### Step 2: Determine the Area of the Semicircle
The area \( A \) of the semicircular wire can be calculated using the formula:
\[ A = \frac{1}{2} \pi r^2 \]

### Step 3: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the semicircular area when it rotates can be expressed as:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where \( \theta \) is the angle between the magnetic field and the normal to the area. As the wire rotates, \( \theta \) changes with time as \( \theta = \omega t \).

Thus, the magnetic flux becomes:
\[ \Phi(t) = B \cdot \left(\frac{1}{2} \pi r^2\right) \cdot \cos(\omega t) \]

### Step 4: Calculate the Induced EMF
According to Faraday's law of electromagnetic induction, the induced EMF \( \mathcal{E} \) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ \mathcal{E} = -\frac{d}{dt}\left(B \cdot \frac{1}{2} \pi r^2 \cos(\omega t)\right) = B \cdot \frac{1}{2} \pi r^2 \cdot \omega \sin(\omega t) \]

### Step 5: Calculate the Current
Using Ohm's law, the current \( I \) in the circuit can be expressed as:
\[ I = \frac{\mathcal{E}}{R} = \frac{B \cdot \frac{1}{2} \pi r^2 \cdot \omega \sin(\omega t)}{R} \]

### Step 6: Determine the Peak Current
The peak current \( I_0 \) can be defined as:
\[ I_0 = \frac{B \cdot \frac{1}{2} \pi r^2 \cdot \omega}{R} \]

### Step 7: Calculate the RMS Current
The root mean square (RMS) current \( I_{rms} \) is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{B \cdot \frac{1}{2} \pi r^2 \cdot \omega}{R \sqrt{2}} \]

### Step 8: Calculate the Mean Power
The mean power \( P \) generated in the circuit is given by:
\[ P = I_{rms}^2 \cdot R \]
Substituting the expression for \( I_{rms} \):
\[ P = \left(\frac{B \cdot \frac{1}{2} \pi r^2 \cdot \omega}{R \sqrt{2}}\right)^2 \cdot R \]
\[ P = \frac{B^2 \cdot \frac{1}{4} \pi^2 r^4 \cdot \omega^2}{2R} \]

### Step 9: Simplify the Expression
This simplifies to:
\[ P = \frac{B^2 \cdot \pi^2 r^4 \cdot \omega^2}{8R} \]

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