The magnitude of the earth's magnetic field at a place is `B_(0)` and angle of dip is `delta.` A horizontal conductor of lenth/lying along the magnetic north-south moves eastwards with a velocity v. The emf induced acroos the coductor is

Options

A. zero
B. `B_(0)lvsindelta`
C. `B_(0)lv`
D. `B_(0)lvcosdelta`

Explanation

To solve the problem, we need to determine the induced electromotive force (emf) across a horizontal conductor that is moving in the Earth's magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
- We have a horizontal conductor of length \( l \) aligned along the magnetic north-south direction.
- The conductor is moving eastwards with a velocity \( v \).
- The Earth's magnetic field at this location has a magnitude \( B_0 \) and an angle of dip \( \delta \).

### Step 2: Identify the Components of the Magnetic Field
- The Earth's magnetic field can be resolved into two components:
- A vertical component \( B_v = B_0 \sin(\delta) \)
- A horizontal component \( B_h = B_0 \cos(\delta) \)

Since the conductor is horizontal and moving east, we focus on the vertical component of the magnetic field for the calculation of induced emf.

### Step 3: Apply Faraday's Law of Electromagnetic Induction
- The induced emf (\( \mathcal{E} \)) in a conductor moving through a magnetic field can be calculated using the formula:
\[ \mathcal{E} = B \cdot l \cdot v \]
where \( B \) is the magnetic field component perpendicular to the direction of motion, \( l \) is the length of the conductor, and \( v \) is the velocity of the conductor.

### Step 4: Substitute the Vertical Component of the Magnetic Field
- In this case, the vertical component of the magnetic field is \( B_v = B_0 \sin(\delta) \).
- Therefore, substituting this into the formula for induced emf gives:
\[ \mathcal{E} = B_v \cdot l \cdot v = (B_0 \sin(\delta)) \cdot l \cdot v \]

### Step 5: Final Expression for Induced EMF
- The induced emf across the conductor is:
\[ \mathcal{E} = B_0 \cdot l \cdot v \cdot \sin(\delta) \]

Thus, the final answer is:
\[ \mathcal{E} = B_0 l v \sin(\delta) \]