At a perpendicular place on the earth, the horizontal component of earth's magnetic field of B, and the angle of dip is `theta`. A striaght meridian and is moved horizontally perpendicular to its length with a velocity v. The emf induced across the rod is

Options

A. `Bvl sintheta`
B. `Bvl cos theta`
C. Bvl tan theta`
D. Bvl

Answer

To solve the problem, we need to find the induced EMF across a straight rod that is moved horizontally in the presence of the Earth's magnetic field. Let's break down the solution step by step.

### Step 1: Understand the components of the Earth's magnetic field
The Earth's magnetic field can be resolved into two components:
- The horizontal component \( H \) (which is given as \( B \))
- The vertical component \( V \)

The angle of dip \( \theta \) relates these two components through the tangent function:
\[ \tan(\theta) = \frac{V}{H} \]

### Step 2: Calculate the vertical component of the magnetic field
From the relationship established in Step 1, we can express the vertical component \( V \) in terms of the horizontal component \( H \) and the angle of dip \( \theta \):
\[ V = H \tan(\theta) \]
Since \( H = B \), we can substitute:
\[ V = B \tan(\theta) \]

### Step 3: Apply the formula for induced EMF
The formula for induced EMF (\( \mathcal{E} \)) when a conductor of length \( L \) moves through a magnetic field is given by:
\[ \mathcal{E} = B \cdot L \cdot v \cdot \sin(\phi) \]
where \( \phi \) is the angle between the direction of motion and the magnetic field.

### Step 4: Determine the angle \( \phi \)
In this case, the rod is moving horizontally, and the vertical component of the magnetic field is acting downwards. Therefore, the angle \( \phi \) between the direction of motion (horizontal) and the vertical magnetic field is \( 90^\circ - \theta \). Thus, we have:
\[ \sin(\phi) = \sin(90^\circ - \theta) = \cos(\theta) \]

### Step 5: Substitute into the EMF formula
Now we can substitute the vertical component \( V \) and the angle \( \phi \) into the induced EMF formula:
\[ \mathcal{E} = V \cdot L \cdot v \cdot \cos(\theta) \]
Substituting \( V = B \tan(\theta) \):
\[ \mathcal{E} = (B \tan(\theta)) \cdot L \cdot v \cdot \cos(\theta) \]

### Step 6: Simplify the expression
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can simplify:
\[ \mathcal{E} = B \cdot L \cdot v \cdot \sin(\theta) \]

### Final Answer
Thus, the induced EMF across the rod is:
\[ \mathcal{E} = B \cdot L \cdot v \cdot \sin(\theta) \]