The magnitude of the earth's magnetic field at the north pole is `B_(0)` . A horizontal conductor of length `l` moves with a velocity `v` . The direction of `v` is perpendicular to the conductor. The induced emf is
(i) zero, if `v` is vertical
(ii) `B_(0)lv` , if `v` is vertical
(iii) zero, if `v` is horizontal
(iv) `B_(0)lv` , if `v` is horizontal

Options

A. zero, if v is vertical
B. `B_(0)`lv, if v is vertical
C. zero, if v is horizontal
D. `B_(0)` lv, if v is horizontal

Answer

To solve the problem, we will analyze the situation based on the given conditions and apply the principles of electromagnetic induction.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a horizontal conductor of length \( l \) moving with a velocity \( v \).
- The earth's magnetic field at the North Pole is denoted as \( B_0 \).
- The direction of the velocity \( v \) is given as perpendicular to the conductor.

2. **Case Analysis**:
- We need to analyze two cases based on the direction of the velocity \( v \): when \( v \) is vertical and when \( v \) is horizontal.

3. **Case 1: \( v \) is Vertical**:
- If \( v \) is vertical, it means that the velocity vector \( v \) is along the z-axis (let's assume \( \hat{k} \)).
- The magnetic field \( B \) at the North Pole is directed downwards (along the negative z-axis).
- Since the direction of \( v \) is vertical and the conductor is horizontal, we can conclude that the velocity vector and the magnetic field vector are parallel.
- The induced EMF \( \mathcal{E} \) can be calculated using the formula:
\[ \mathcal{E} = l \cdot (v \times B) \]
- Since \( v \) is parallel to \( B \), the cross product \( v \times B \) will be zero.
- Thus, the induced EMF is:
\[ \mathcal{E} = 0 \]

4. **Case 2: \( v \) is Horizontal**:
- If \( v \) is horizontal, we can assume \( v \) is along the x-axis (let's say \( \hat{i} \)).
- The magnetic field \( B \) is still directed downwards (along the negative z-axis).
- The conductor is horizontal, and we can assume it is along the y-axis (let's say \( \hat{j} \)).
- Now, we can calculate the induced EMF using the same formula:
\[ \mathcal{E} = l \cdot (v \times B) \]
- Here, \( v \) is along \( \hat{i} \) and \( B \) is along \( -\hat{k} \).
- The cross product \( v \times B \) can be calculated:
\[ \hat{i} \times (-\hat{k}) = \hat{j} \]
- Therefore, the induced EMF becomes:
\[ \mathcal{E} = l \cdot (v \cdot \hat{j}) = l \cdot v \]
- Since the magnitude of the magnetic field is \( B_0 \), we can express the induced EMF as:
\[ \mathcal{E} = B_0 \cdot l \cdot v \]

5. **Conclusion**:
- From the analysis, we conclude:
- The induced EMF is **zero** if \( v \) is vertical.
- The induced EMF is \( B_0 l v \) if \( v \) is horizontal.

### Final Answer:
- (i) Zero, if \( v \) is vertical.
- (iv) \( B_0 l v \), if \( v \) is horizontal.