Question

A metal conductor of length $1\ \text{m}$ rotates vertically about one of its ends at an angular velocity of $5\ \text{rad/s}$. If the horizontal component of earth's magnetic field is $0.2 \times 10^{-4}\ \text{T}$, then the emf developed between the two ends of the conductor is

• A. $5\ \mu\text{V}$
• B. $50\ \mu\text{V}$
• C. $25\ \mu\text{V}$
• D. $100\ \mu\text{V}$

Hint:

For a rotating vertical rod, it cuts horizontal magnetic field lines, so you must use $B_H$. If it were rotating horizontally (like a sweeping radar), it would cut vertical field lines, requiring $B_V$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A metal rod rotates about one of its ends in the Earth's magnetic field. We need to find the motional electromotive force (emf) induced across its ends.

Step 2: Key Formula or Approach:
The motional emf $e$ induced in a conducting rod of length $L$ rotating with an angular velocity $\omega$ in a uniform perpendicular magnetic field $B$ is given by the formula:
$$e = \frac{1}{2} B \omega L^2$$

Step 3: Detailed Explanation:
Let's list the given parameters:
Length of the conductor, $L = 1\ \text{m}$
Angular velocity, $\omega = 5\ \text{rad/s}$
Horizontal component of Earth's magnetic field, $B = 0.2 \times 10^{-4}\ \text{T}$
As the rod rotates vertically, it sweeps perpendicularly through the horizontal component of the magnetic field.
Substitute these values into the formula:
$$e = \frac{1}{2} \times (0.2 \times 10^{-4}) \times 5 \times (1)^2$$ $$e = 0.1 \times 10^{-4} \times 5$$ $$e = 0.5 \times 10^{-4}\ \text{V}$$ To convert this value to microvolts ($\mu\text{V}$), we multiply by $10^6$:
$$e = 0.5 \times 10^{-4} \times 10^6\ \mu\text{V}$$ $$e = 0.5 \times 10^2\ \mu\text{V} = 50\ \mu\text{V}$$

Step 4: Final Answer:
The developed emf is $50\ \mu\text{V}$, matching option (B).