Question:
In an a.c. generator, a rectangular coil of \(1000\) turns each having area \(\frac{10^{-2}}{\pi }\ \text{m}^{2}\) is rotated at \(180\) r.p.m. about an axis perpendicular to a uniform magnetic field of magnitude \(2\) T. The maximum value of e.m.f. produced will be
In an a.c. generator, a rectangular coil of \(1000\) turns each having area \(\frac{10^{-2}}{\pi }\ \text{m}^{2}\) is rotated at \(180\) r.p.m. about an axis perpendicular to a uniform magnetic field of magnitude \(2\) T. The maximum value of e.m.f. produced will be
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Peak emf is n B A omega, with omega from the rotation rate in revolutions per second.
Updated On: Nov 12, 2026
- \(60\) V
- \(120\) V
- \(180\) V
- \(200\) V
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Peak emf \(\epsilon_0=nBA\omega\).
Step 2: Angular speed
\(180\) r.p.m. \(=3\) rev/s, so \(\omega=2\pi\times3=6\pi\) rad/s.
Step 3: Calculation
\(\epsilon_0=1000\times2\times\dfrac{10^{-2}}{\pi}\times6\pi=120\) V
Final Answer:
The maximum emf is 120 V. \[ \boxed{\text{(B) }120\ \text{V}} \]
Peak emf \(\epsilon_0=nBA\omega\).
Step 2: Angular speed
\(180\) r.p.m. \(=3\) rev/s, so \(\omega=2\pi\times3=6\pi\) rad/s.
Step 3: Calculation
\(\epsilon_0=1000\times2\times\dfrac{10^{-2}}{\pi}\times6\pi=120\) V
Final Answer:
The maximum emf is 120 V. \[ \boxed{\text{(B) }120\ \text{V}} \]
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