Consider a variation of the previous problem. Suppose the circular loop lies in a vertical plane. The rod has a mass m. The rod and the loop have negligible resistance but the wire connecting O and C has a resistance R. The rod is made to rotate with a uniform angular velocity `(omega)` in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it.
Find the magnitude of this force when the rod makes an angle `(theta)` with the vertical.

Options

A. `(B^(2)a^(3)omega)/(2R)-mg sin theta`
B. `(B^(2)a^(3)omega)/(2R)+mg sin theta`
C. `(B^(2)a^(3)omega)/(R)-mg sin theta`
D. None of these

Explanation

To solve the problem, we need to analyze the forces acting on the rod and the induced electromotive force (EMF) due to its rotation in a magnetic field. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Rod
The rod has a weight \( mg \) acting downwards. When the rod makes an angle \( \theta \) with the vertical, the weight can be resolved into two components:
- The component acting along the direction of the rod: \( mg \cos \theta \)
- The component acting perpendicular to the rod: \( mg \sin \theta \)

### Step 2: Determine the Induced EMF
As the rod rotates in a magnetic field, an electromotive force (EMF) is induced in it. The formula for the induced EMF \( E \) when a rod of length \( A \) rotates with angular velocity \( \omega \) in a magnetic field \( B \) is given by:
\[ E = \frac{1}{2} B \omega A^2 \]

### Step 3: Calculate the Current
The current \( I \) flowing through the circuit can be calculated using Ohm's law:
\[ I = \frac{E}{R} = \frac{\frac{1}{2} B \omega A^2}{R} = \frac{B \omega A^2}{2R} \]

### Step 4: Calculate the Magnetic Force
The magnetic force \( F_B \) acting on the rod due to the current in the magnetic field is given by:
\[ F_B = I \cdot B \cdot L = \left( \frac{B \omega A^2}{2R} \right) \cdot B \cdot A = \frac{B^2 \omega A^3}{2R} \]

### Step 5: Set Up the Force Balance
To maintain a constant angular velocity, the net force acting on the rod must be zero. The magnetic force \( F_B \) acts in one direction, while the component of the weight \( mg \sin \theta \) acts in the opposite direction. Thus, we can write the force balance as:
\[ F_B - mg \sin \theta = 0 \]
This implies:
\[ F_B = mg \sin \theta \]

### Step 6: Substitute for \( F_B \)
Substituting the expression for \( F_B \) into the force balance equation gives:
\[ \frac{B^2 \omega A^3}{2R} = mg \sin \theta \]

### Step 7: Solve for the Force \( F \)
Now, we can express the force \( F \) that needs to be applied at the midpoint of the rod to maintain the rotation:
\[ F = mg \sin \theta - \frac{B^2 \omega A^3}{2R} \]

### Final Answer
Thus, the magnitude of the force \( F \) required to maintain the uniform angular velocity \( \omega \) when the rod makes an angle \( \theta \) with the vertical is:
\[ F = mg \sin \theta - \frac{B^2 \omega A^3}{2R} \]