A rod of length L, along east-west direction is dropped from a height H. If B be the magnetic field due to Earth at that place and angle of dip is `theta,` then the magnitude of the induced e.m.f. across two ends of the rod when the rod reachs the Earth is

Options

A. `BLH cos theta`
B. `BL cos theta xx (2gH) ^(1//2)`
C. `BL cos theta //(2 gH)^(1//2)`
D. None of the above

Explanation

To find the magnitude of the induced e.m.f. across the ends of the rod when it reaches the Earth, we can follow these steps:

### Step 1: Understand the scenario
A rod of length \( L \) is dropped from a height \( H \) in a magnetic field \( B \) with an angle of dip \( \theta \). The rod is oriented along the east-west direction.

**Hint:** Visualize the rod falling and the magnetic field lines acting on it.

### Step 2: Determine the final velocity of the rod
When the rod is dropped from height \( H \), it will fall under the influence of gravity. The final velocity \( v \) of the rod just before it reaches the ground can be calculated using the equation of motion:
\[ v = \sqrt{2gH} \]
where \( g \) is the acceleration due to gravity.

**Hint:** Use the equation of motion that relates initial velocity, final velocity, acceleration, and distance.

### Step 3: Find the horizontal component of the magnetic field
The magnetic field \( B \) has a component in the direction of the rod's motion. Since the angle of dip is \( \theta \), the horizontal component of the magnetic field \( B_h \) is given by:
\[ B_h = B \cos \theta \]

**Hint:** Remember that the angle of dip affects the orientation of the magnetic field.

### Step 4: Calculate the induced e.m.f.
The induced e.m.f. \( \mathcal{E} \) across the ends of the rod can be calculated using the formula:
\[ \mathcal{E} = B_h \cdot L \cdot v \]
Substituting the values we have:
\[ \mathcal{E} = (B \cos \theta) \cdot L \cdot \sqrt{2gH} \]

**Hint:** The induced e.m.f. is proportional to the length of the rod, the magnetic field component, and the velocity of the rod.

### Step 5: Final expression for induced e.m.f.
Thus, the magnitude of the induced e.m.f. across the two ends of the rod when it reaches the Earth is:
\[ \mathcal{E} = B L \cos \theta \sqrt{2gH} \]

**Hint:** Ensure to keep track of the units and the physical significance of each term in the equation.

### Conclusion
The final answer is:
\[ \mathcal{E} = B L \cos \theta \sqrt{2gH} \]