Question

A straight metallic rod of length $L$ is oriented horizontally and dropped vertically downward with a speed $v$ perpendicular to the horizontal component of the Earth's uniform magnetic field $B_H$. The instantaneous motional electromotive force (EMF) induced across the two ends of this rod is:

• A. \( \frac{1}{2}B_H v L^2 \)
• B. \( B_H v L \)
• C. \( \frac{B_H v}{L} \)
• D. \( 0 \)

Hint:

For motional EMF to exist, the moving conductor must actively cut across magnetic flux lines. If you align the rod along the North-South direction and drop it, its length axis runs parallel to the field lines, which means zero lines are cut and the induced EMF drops to exactly zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Motional electromotive force (EMF) is a potential difference that arises across a conductor when it moves through a magnetic field region in a way that "cuts" across the magnetic flux lines.
This phenomenon can be explained by the Lorentz force acting on the free conduction electrons within the metal.
As the rod moves with velocity \(\vec{v}\), every electron inside it also moves with velocity \(\vec{v}\).
Since these electrons are moving through an external magnetic field \(\vec{B}\), they experience a force \(\vec{F} = -e(\vec{v} \times \vec{B})\).
This force drives electrons toward one end of the rod, leaving a net positive charge at the other end.
This separation of charge creates an internal electric field that eventually balances the magnetic force, resulting in a stable voltage (EMF) across the rod's length.
Step 2: Key Formula or Approach:
The general expression for the motional EMF \(e\) induced in a straight conductor is given by the scalar triple product:
\[ e = (\vec{v} \times \vec{B}) \cdot \vec{L} \]
Where:
- \(\vec{v}\) is the velocity of the conductor.
- \(\vec{B}\) is the magnetic field intensity.
- \(\vec{L}\) is the vector representing the length and orientation of the conductor.
Step 3: Detailed Explanation:
Let's analyze the geometry described in the question:
1. Rod Orientation: The rod is oriented horizontally. Let's assume its length vector \(\vec{L}\) points along the East-West direction.
2. Motion: The rod is dropped vertically downward. Therefore, its velocity vector \(\vec{v}\) points straight down toward the Earth's surface.
3. Magnetic Field: The field involved is the horizontal component of Earth's magnetic field (\(B_H\)), which typically points from magnetic South to North.
In this setup:
- \(\vec{v}\) (Vertical) is perpendicular to \(\vec{B}_H\) (Horizontal N-S).
- The resulting cross product \((\vec{v} \times \vec{B}_H)\) points in a horizontal direction (East or West, depending on the specific hemisphere).
- The rod's length \(\vec{L}\) (Horizontal E-W) is parallel to the direction of the cross product vector.
Since all three vectors (\(\vec{v}, \vec{B}_H, \vec{L}\)) are mutually perpendicular, the sine and cosine terms in the triple product evaluate to 1:
\[ e = |v| \cdot |B_H| \cdot \sin(90^{\circ}) \cdot |L| \cdot \cos(0^{\circ}) \]
\[ e = v \cdot B_H \cdot 1 \cdot L \cdot 1 \]
\[ e = B_H v L \]
This configuration represents the most efficient way to generate motional EMF, as the rod perfectly cuts across the magnetic field lines.
Step 4: Final Answer:
The induced instantaneous motional EMF is \(B_H v L\).