Question

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

• A. \(5\mu V\)
• B. \(5\) mV
• C. \(50\mu V\)
• D. \(50\) mV

Hint:

The emf across a rod rotating about one end is \(\dfrac{1}{2}B\omega l^{2}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
A rod rotating about one end in a perpendicular field has an emf \(\varepsilon=\dfrac{1}{2}B\omega l^{2}\).

Step 2: Substitute
\(B=0.2\times 10^{-4}\) T, \(\omega=5\) rad/s, \(l=1\) m

Step 3: Calculation
\(\varepsilon=\dfrac{1}{2}\times 0.2\times 10^{-4}\times 5\times 1^{2}=5\times 10^{-5}\) V
\(\varepsilon=50\ \mu\text{V}\)

Step 4: Compare the options
The emf is \(50\,\mu\)V, option (C).

Final Answer:
The emf developed is 50 microvolt. \[ \boxed{\text{(C) }50\,\mu\text{V}} \]