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: Picture the rod.
A rod of length $1$ m spins about one end in the Earth's magnetic field at $\omega=5\ \text{rad/s}$. We want the emf between its two ends.

Step 2: Recall the spinning-rod emf formula.
For a rod rotating about one end in a field $B$ perpendicular to its sweep, \[ e=\frac{1}{2}B\omega L^2. \]

Step 3: Why the half and the $L^2$.
Different parts of the rod move at different speeds. The far end moves fastest. Averaging over the whole rod brings the factor $\frac{1}{2}$ and the $L^2$.

Step 4: List the values.
$L=1$ m, $\omega=5\ \text{rad/s}$, $B=0.2\times10^{-4}\ \text{T}$ (the horizontal part the vertical rod sweeps through).

Step 5: Substitute into the formula.
\[ e=\frac{1}{2}\times(0.2\times10^{-4})\times5\times(1)^2. \]

Step 6: Do the arithmetic.
\[ e=0.1\times10^{-4}\times5=0.5\times10^{-4}\ \text{V}. \]

Step 7: Convert to microvolts.
Multiply by $10^6$: $0.5\times10^{-4}\times10^6=50\ \mu\text{V}$. This is option (2).
\[ \boxed{50\ \mu\text{V}} \]