The magnetic field in a region is given by `vec(B)=B_(0)(1+(x)/(a))hat(k)`. A square loop of edge length 'd' is placed with its edge along X-axis and Y-axis. The loop is moved with a constant velocity `vec(V)=V_(0)hat(i)`. The emf induced in the loop is

Options

A. `(v_(0)B_(0)d^(2))/(a)`
B. `(v_(0)B_(0)d^(3))/(a)`
C. `(v_(0)B_(0)d^(2))/(2a)`
D. `0`

Explanation

To solve the problem of finding the induced EMF in the square loop moving in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnetic Field**:
The magnetic field is given by:
\[ \vec{B} = B_0 \left(1 + \frac{x}{a}\right) \hat{k} \]
This indicates that the magnetic field varies with the position \(x\).

2. **Define the Loop's Position**:
The square loop has an edge length \(d\) and is placed in the xy-plane, with its edges along the x-axis and y-axis. The loop is moving with a constant velocity \(\vec{V} = V_0 \hat{i}\).

3. **Calculate the Magnetic Flux**:
The magnetic flux \(\Phi\) through the loop is given by:
\[ \Phi = \int \vec{B} \cdot d\vec{A} \]
For a small strip of the loop of thickness \(dx\) at position \(x\), the area \(dA\) is \(d \cdot dx\) (where \(d\) is the length of the edge of the square loop). Thus, we can write:
\[ d\Phi = B \cdot dA = B_0 \left(1 + \frac{x}{a}\right) d \cdot dx \]

4. **Integrate to Find Total Flux**:
The total flux through the loop as it moves from \(x\) to \(x + d\) is:
\[ \Phi = \int_{x}^{x+d} B_0 \left(1 + \frac{x}{a}\right) d \cdot dx \]
This can be simplified as:
\[ \Phi = B_0 d \int_{x}^{x+d} \left(1 + \frac{x}{a}\right) dx \]
Evaluating this integral gives:
\[ \Phi = B_0 d \left[ x + \frac{x^2}{2a} \right]_{x}^{x+d} = B_0 d \left[ \left(x+d + \frac{(x+d)^2}{2a}\right) - \left(x + \frac{x^2}{2a}\right) \right] \]

5. **Differentiate to Find EMF**:
According to Faraday's law of electromagnetic induction, the induced EMF (\(\mathcal{E}\)) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Since the loop is moving with velocity \(V_0\), we can express this as:
\[ \mathcal{E} = -\frac{d\Phi}{dx} \cdot \frac{dx}{dt} = -\frac{d\Phi}{dx} \cdot V_0 \]

6. **Calculate the Derivative of Flux**:
After calculating the flux, we differentiate it with respect to \(x\) and multiply by \(V_0\) to find the induced EMF:
\[ \mathcal{E} = B_0 d \cdot \frac{d}{dx} \left( \frac{d^2}{2a} + d \right) \cdot V_0 \]

7. **Final Result**:
After performing the calculations, we find:
\[ \mathcal{E} = \frac{B_0 d^2 V_0}{a} \]

### Final Answer:
The induced EMF in the loop is:
\[ \mathcal{E} = \frac{B_0 d^2 V_0}{a} \]