The magnetic field in a region is given by `vec B = (B_0)/(L)y hatk` where L is a fixed length. A conducting rod of length L lies along the Y - axis between the origin and the point (0,L, 0). If the rod moves with a velocity `v= v_0 hati` , find the `EMF` induced between the ends of the rod.

Options

A. `2 B_(0)v_(0)l`
B. `B_(0)v_(0)l`
C. `(B_(0)v_(0)l)/(2)`
D. None of these

Explanation

To solve the problem of finding the induced EMF in the conducting rod, we can follow these steps:

### Step 1: Understand the Given Information
We have a magnetic field given by:
\[ \vec{B} = \frac{B_0}{L} y \hat{k} \]
where \(B_0\) is a constant, \(L\) is a fixed length, and \(y\) is the position along the y-axis. The conducting rod of length \(L\) lies along the y-axis from the origin (0,0,0) to the point (0,L,0). The rod is moving with a velocity:
\[ \vec{v} = v_0 \hat{i} \]

### Step 2: Identify the Induced EMF Formula
The induced EMF (\(\epsilon\)) in a moving conductor in a magnetic field can be calculated using the formula:
\[ \epsilon = \int \vec{E} \cdot d\vec{l} \]
where \(\vec{E}\) is the electric field induced due to the motion of the rod in the magnetic field, and \(d\vec{l}\) is the differential length element along the rod.

### Step 3: Calculate the Induced EMF for an Element of the Rod
Consider a small element of the rod at a distance \(y\) from the origin with a width \(dy\). The induced EMF across this element can be expressed as:
\[ d\epsilon = B \cdot v \cdot dy \]
Substituting the values:
\[ d\epsilon = \left(\frac{B_0}{L} y\right) \cdot v_0 \cdot dy \]

### Step 4: Integrate to Find Total EMF
Now, we need to integrate this expression from \(y = 0\) to \(y = L\):
\[ \epsilon = \int_0^L d\epsilon = \int_0^L \left(\frac{B_0}{L} y v_0\right) dy \]
This simplifies to:
\[ \epsilon = \frac{B_0 v_0}{L} \int_0^L y \, dy \]

### Step 5: Solve the Integral
The integral \(\int_0^L y \, dy\) can be calculated as:
\[ \int_0^L y \, dy = \left[\frac{y^2}{2}\right]_0^L = \frac{L^2}{2} \]

### Step 6: Substitute Back to Find EMF
Substituting the result of the integral back into the EMF expression:
\[ \epsilon = \frac{B_0 v_0}{L} \cdot \frac{L^2}{2} = \frac{B_0 v_0 L}{2} \]

### Conclusion
Thus, the induced EMF between the ends of the rod is:
\[ \epsilon = \frac{B_0 v_0 L}{2} \]