A metallic rod of length L rotates at an angular speed `omega,` normal to a uniform magnetic field. If the resitance of rod is R the heat dissipated is:

Options

A. `(1)/(4) ( B ^(2) L ^(4) omega ^(2) t)/( R)`
B. `(1)/(2) (B ^(2) L ^(2) omega t)/(R)`
C. `(1 )/(4) (B ^(2) L ^(2) omega ^(2) t ^(2))/( R)`
D. `(1)/(4) (BL^(2) omega ^(4) t)/(R)`

Explanation

To solve the problem, we need to find the heat dissipated in a metallic rod of length \( L \) rotating with an angular speed \( \omega \) in a uniform magnetic field \( B \) with resistance \( R \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a metallic rod of length \( L \) rotating about an axis with angular speed \( \omega \).
- The rod is placed in a uniform magnetic field \( B \), which is perpendicular to the plane of rotation.

2. **Induced EMF Calculation**:
- The induced electromotive force (EMF) in the rod can be calculated using the formula:
\[ E = B \cdot v \cdot L \]
- Here, \( v \) is the linear velocity of a point at a distance \( x \) from the axis of rotation, given by:
\[ v = \omega \cdot x \]
- Therefore, the induced EMF for a small segment \( dx \) of the rod at distance \( x \) is:
\[ dE = B \cdot (\omega \cdot x) \cdot dx \]

3. **Total Induced EMF**:
- To find the total induced EMF \( E \) in the entire rod, we integrate \( dE \) from \( 0 \) to \( L \):
\[ E = \int_0^L B \cdot (\omega \cdot x) \, dx = B \omega \int_0^L x \, dx \]
- The integral of \( x \) from \( 0 \) to \( L \) is:
\[ \int_0^L x \, dx = \frac{L^2}{2} \]
- Thus, the total induced EMF is:
\[ E = B \omega \cdot \frac{L^2}{2} = \frac{B \omega L^2}{2} \]

4. **Current Calculation**:
- Using Ohm's law, the current \( I \) flowing through the rod can be calculated as:
\[ I = \frac{E}{R} = \frac{\frac{B \omega L^2}{2}}{R} = \frac{B \omega L^2}{2R} \]

5. **Heat Dissipated**:
- The heat \( Q \) dissipated in the rod over a time \( T \) can be calculated using the formula:
\[ Q = I^2 R T \]
- Substituting for \( I \):
\[ Q = \left( \frac{B \omega L^2}{2R} \right)^2 R T \]
- Simplifying this expression:
\[ Q = \frac{B^2 \omega^2 L^4}{4R^2} R T = \frac{B^2 \omega^2 L^4 T}{4R} \]

### Final Answer:
The heat dissipated in the rod is:
\[ Q = \frac{B^2 \omega^2 L^4 T}{4R} \]