A conductor `AB` of length `l` moves in `x y` plane with velocity `vec(v) = v_(0)(hat(i)-hat(j))`. A magnetic field `vec(B) = B_(0) (hat(i) + hat(j))` exists in the region. The induced emf is

Options

A. (a) zero
B. (b) `2B_(0)lv_(0)`
C. (c ) `B_(0)lv_(0)`
D. (d) `sqer(2)B_(0)lv_(0)`

Explanation

To find the induced EMF in the given scenario, we can use the formula for induced EMF, which is given by:

\[ \text{EMF} = \vec{v} \times \vec{B} \cdot \vec{l} \]

where:
- \(\vec{v}\) is the velocity vector of the conductor,
- \(\vec{B}\) is the magnetic field vector,
- \(\vec{l}\) is the length vector of the conductor.

### Step 1: Identify the vectors
Given:
- The velocity vector \(\vec{v} = v_0 (\hat{i} - \hat{j})\)
- The magnetic field vector \(\vec{B} = B_0 (\hat{i} + \hat{j})\)

### Step 2: Set up the cross product
We need to compute the cross product \(\vec{v} \times \vec{B}\). We can represent this using the determinant of a matrix:

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_0 & -v_0 & 0 \\ B_0 & B_0 & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:

\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} -v_0 & 0 \\ B_0 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} v_0 & 0 \\ B_0 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} v_0 & -v_0 \\ B_0 & B_0 \end{vmatrix} \]

Calculating each of these determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} -v_0 & 0 \\ B_0 & 0 \end{vmatrix} = (-v_0)(0) - (0)(B_0) = 0 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} v_0 & 0 \\ B_0 & 0 \end{vmatrix} = (v_0)(0) - (0)(B_0) = 0 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} v_0 & -v_0 \\ B_0 & B_0 \end{vmatrix} = (v_0)(B_0) - (-v_0)(B_0) = v_0 B_0 + v_0 B_0 = 2v_0 B_0 \]

Thus, we have:

\[ \vec{v} \times \vec{B} = 0 \hat{i} - 0 \hat{j} + 2v_0 B_0 \hat{k} = 2v_0 B_0 \hat{k} \]

### Step 4: Dot product with length vector
Now, we need to dot this result with the length vector \(\vec{l}\). Since \(\vec{l}\) is in the xy-plane, we can express it as:

\[ \vec{l} = l_x \hat{i} + l_y \hat{j} \]

The dot product is:

\[ (2v_0 B_0 \hat{k}) \cdot (l_x \hat{i} + l_y \hat{j}) = 0 \]

### Step 5: Conclusion
Since the dot product is zero, the induced EMF is:

\[ \text{EMF} = 0 \]

### Final Answer
The induced EMF is \(0\).

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