A wire of length 4 m placed normal to the plane magnetic field of `(2 hat (i) + 4 hat (j))` Tesla is moving with a velocity `(4 hat (i)+ 6 hat (j)+ 8 hat (k))` m/s. The emf induced across the ends of the wire will be __

Options

A. 4V
B. 8V
C. 16V
D. 32V

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Length of the wire, \( L = 4 \, \text{m} \)
- Magnetic field, \( \mathbf{B} = 2 \hat{i} + 4 \hat{j} \, \text{T} \)
- Velocity of the wire, \( \mathbf{v} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \, \text{m/s} \)

2. **Determine the Direction of the Wire**:
- The wire is placed normal to the plane of the magnetic field. The magnetic field lies in the \( \hat{i} \) and \( \hat{j} \) directions, so the wire is aligned along the \( \hat{k} \) direction.
- Therefore, we can represent the length vector of the wire as:
\[ \mathbf{L} = 4 \hat{k} \, \text{m} \]

3. **Use the Formula for Induced EMF**:
- The induced EMF (\( \mathcal{E} \)) can be calculated using the formula:
\[ \mathcal{E} = \mathbf{L} \cdot (\mathbf{v} \times \mathbf{B}) \]

4. **Calculate the Cross Product \( \mathbf{v} \times \mathbf{B} \)**:
- First, we need to compute the cross product \( \mathbf{v} \times \mathbf{B} \):
\[ \mathbf{v} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \]
\[ \mathbf{B} = 2 \hat{i} + 4 \hat{j} \]
- Using the determinant method for the cross product:
\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 8 \\ 2 & 4 & 0 \end{vmatrix} \]
- Expanding the determinant:
\[ = \hat{i}(6 \cdot 0 - 8 \cdot 4) - \hat{j}(4 \cdot 0 - 8 \cdot 2) + \hat{k}(4 \cdot 4 - 6 \cdot 2) \]
\[ = \hat{i}(0 - 32) - \hat{j}(0 - 16) + \hat{k}(16 - 12) \]
\[ = -32 \hat{i} + 16 \hat{j} + 4 \hat{k} \]

5. **Calculate the Dot Product**:
- Now, we calculate \( \mathbf{L} \cdot (\mathbf{v} \times \mathbf{B}) \):
\[ \mathcal{E} = \mathbf{L} \cdot (-32 \hat{i} + 16 \hat{j} + 4 \hat{k}) \]
\[ = 4 \hat{k} \cdot (-32 \hat{i} + 16 \hat{j} + 4 \hat{k}) = 4 \cdot 4 = 16 \, \text{V} \]

6. **Final Result**:
- The induced EMF across the ends of the wire is:
\[ \mathcal{E} = 16 \, \text{V} \]