The magnetic field in a region is given by `vecB=B_(0)/Lxhatk` where `L` is a fixed length.A conducting rod of length lies along the `X`-axis between the origin and the point `(L,0,0)`.If the rod moves with a velocity `vecv=v_(0) hatj` the `emf` induced between the ends of the rod.

Answer

To solve the problem, we need to find the electromotive force (emf) induced in a conducting rod that is moving in a magnetic field. Let's break down the solution step by step.

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

### Step 2: Identify the Induced EMF Formula
The induced emf (\(E\)) in a moving conductor in a magnetic field can be calculated using the formula:
\[ E = B \cdot L \cdot v \]
where \(B\) is the magnetic field, \(L\) is the length of the rod, and \(v\) is the velocity of the rod.

### Step 3: Determine the Magnetic Field at the Position of the Rod
Since the magnetic field depends on the x-coordinate of the rod, we can express the magnetic field at a distance \(x\) along the rod as:
\[ B(x) = \frac{B_0}{L} x \hat{k} \]

### Step 4: Calculate the Induced EMF
To find the total emf induced across the length of the rod, we can consider a small segment \(dx\) of the rod at position \(x\). The induced emf for this segment is given by:
\[ dE = B(x) \cdot v \cdot dx \]
Substituting \(B(x)\) and \(v\):
\[ dE = \left(\frac{B_0}{L} x\right) v_0 \, dx \]

### Step 5: Integrate to Find Total EMF
To find the total emf across the entire length of the rod, we integrate from \(x = 0\) to \(x = L\):
\[ E = \int_0^L \left(\frac{B_0}{L} x v_0\right) dx \]
This simplifies to:
\[ E = \frac{B_0 v_0}{L} \int_0^L x \, dx \]
Calculating the integral:
\[ \int_0^L x \, dx = \frac{x^2}{2} \bigg|_0^L = \frac{L^2}{2} \]
Thus, substituting back:
\[ E = \frac{B_0 v_0}{L} \cdot \frac{L^2}{2} = \frac{B_0 v_0 L}{2} \]

### Final Result
The induced emf between the ends of the rod is:
\[ E = \frac{B_0 v_0 L}{2} \]