A copper rod of length l is rotated about one end perpendicular to the uniform magnetic field B with constant angular velocity `omega` . The induced e.m.f. between its two ends is

Options

A. `Bomegal^(2)`
B. `(3)/(2)Bomegal^(2)`
C. `(1)/(2)Bomegal^(2)`
D. `2 Bomegal^(2)`

Answer

To find the induced electromotive force (e.m.f.) between the two ends of a rotating copper rod in a magnetic field, we can follow these steps:

### Step 1: Understand the Setup
We have a copper rod of length \( L \) that is rotating about one of its ends with a constant angular velocity \( \omega \). The magnetic field \( B \) is uniform and perpendicular to the plane of rotation of the rod.

### Step 2: Determine the Linear Velocity
The linear velocity \( v \) of a point on the rod at a distance \( r \) from the pivot (the end about which it is rotating) is given by:
\[ v = \omega r \]
For a point at the end of the rod, \( r = L \), so:
\[ v = \omega L \]

### Step 3: Calculate the Velocity at the Midpoint
To find the average velocity of the rod, we can consider the midpoint of the rod, which is at \( r = \frac{L}{2} \):
\[ v_{\text{mid}} = \omega \left(\frac{L}{2}\right) = \frac{\omega L}{2} \]

### Step 4: Use the Formula for Induced EMF
The induced e.m.f. \( \mathcal{E} \) can be calculated using the formula:
\[ \mathcal{E} = v \cdot B \cdot L \]
Since \( v \) is the average velocity of the rod, we substitute \( v_{\text{mid}} \):
\[ \mathcal{E} = \left(\frac{\omega L}{2}\right) \cdot B \cdot L \]

### Step 5: Simplify the Expression
Now, simplifying the expression:
\[ \mathcal{E} = \frac{\omega L}{2} \cdot B \cdot L = \frac{1}{2} B \omega L^2 \]

### Final Answer
Thus, the induced e.m.f. between the two ends of the rod is:
\[ \mathcal{E} = \frac{1}{2} B \omega L^2 \]