A uniform magnetic field existsin region given by `vec(B) = 3 hat(i) + 4 hat(j)+5hat(k)`. A rod of length `5 m` is placed along `y`-axis is moved along `x`- axis with constant speed `1 m//sec`. Then the magnitude of induced `e.m.f` in the rod is :

Options

A. zero
B. 25V
C. 20V
D. 15V

Answer

To solve the problem, we need to find the magnitude of the induced electromotive force (e.m.f) in a rod moving through a magnetic field. Here are the steps to arrive at the solution:

### Step 1: Identify the given quantities
- The magnetic field vector is given as:
\[ \vec{B} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \, \text{T} \]
- The length of the rod is:
\[ L = 5 \, \text{m} \]
- The rod is placed along the y-axis, so its vector representation is:
\[ \vec{L} = 5 \hat{j} \, \text{m} \]
- The speed of the rod moving along the x-axis is:
\[ \vec{v} = 1 \hat{i} \, \text{m/s} \]

### Step 2: Use the formula for induced e.m.f
The induced e.m.f (\( \mathcal{E} \)) in the rod can be calculated using the formula:
\[ \mathcal{E} = \vec{L} \cdot (\vec{v} \times \vec{B}) \]

### Step 3: Calculate the cross product \( \vec{v} \times \vec{B} \)
First, we need to compute the cross product \( \vec{v} \times \vec{B} \):
\[ \vec{v} = 1 \hat{i} \]
\[ \vec{B} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \]

Using the determinant method for the cross product:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 3 & 4 & 5 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i}(0 \cdot 5 - 0 \cdot 4) - \hat{j}(1 \cdot 5 - 0 \cdot 3) + \hat{k}(1 \cdot 4 - 0 \cdot 3) \]
\[ = 0 \hat{i} - 5 \hat{j} + 4 \hat{k} \]
\[ = -5 \hat{j} + 4 \hat{k} \]

### Step 4: Calculate the dot product \( \vec{L} \cdot (\vec{v} \times \vec{B}) \)
Now, we compute the dot product:
\[ \vec{L} = 5 \hat{j} \]
\[ \vec{v} \times \vec{B} = -5 \hat{j} + 4 \hat{k} \]
\[ \vec{L} \cdot (\vec{v} \times \vec{B}) = (5 \hat{j}) \cdot (-5 \hat{j} + 4 \hat{k}) \]
Calculating the dot product:
\[ = 5 \cdot (-5) + 0 = -25 \]

### Step 5: Find the magnitude of the induced e.m.f
The magnitude of the induced e.m.f is:
\[ |\mathcal{E}| = |-25| = 25 \, \text{V} \]

### Final Answer
Thus, the magnitude of the induced e.m.f in the rod is:
\[ \mathcal{E} = 25 \, \text{V} \]

---