Flux Linkage In A Rotating Coil
A rectangular coil of 200 turns and area 0.05 m^2 rotates uniformly in a 0.4 T magnetic field, completing 50 revolutions each second. What is the peak EMF generated by this AC generator coil?
Select the correct option:
Solution
\(1256\) V
A coil rotating in a uniform field experiences a sinusoidally varying flux, and Faraday's law gives a peak EMF (\varepsilon_0 = NBA\omega), where (\omega) is the angular speed of rotation. This is the operating principle of an AC generator. The frequency is 50 rev/s, so (\omega = 2\pi \times 50 = 100\pi \approx 314) rad/s. Substituting (N = 200), (B = 0.4) T, (A = 0.05) m^2 and (\omega = 314) rad/s gives (\varepsilon_0 = 200 \times 0.4 \times 0.05 \times 314 = 4 \times 314 = 1256) V. The option 628 V halves the result by using (\pi\times50) instead of (2\pi\times50). The option 400 V drops the angular-speed factor and keeps only (NBA) scaled. The option 200 V uses the turns count alone. This is the NCERT generator relation in which peak EMF scales with rotation rate, arising because the flux varies as (\Phi = NBA\cos\omega t) and its time derivative introduces the factor (\omega). The instantaneous EMF is therefore (\varepsilon = NBA\omega\sin\omega t), a pure sinusoid whose maximum is the peak value computed above and whose RMS value would be smaller by a factor of (\sqrt{2}). A plausibility check confirms the units (T·m^2·rad/s per turn, times turns) give volts, and a high peak is entirely reasonable for a many-turn coil spinning rapidly in a moderate field, which is why practical generators use many turns and high rotation speeds.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic induction and alternating currents
- Topic
- flux linkage in a rotating coil
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
\(1256\) V
A coil rotating in a uniform field experiences a sinusoidally varying flux, and Faraday's law gives a peak EMF (\varepsilon_0 = NBA\omega), where (\omega) is the angular speed of rotation. This is the operating principle of an AC generator. The frequency is 50 rev/s, so (\omega = 2\pi \times 50 = 100\pi \approx 314) rad/s. Substituting (N = 200), (B = 0.4) T, (A = 0.05) m^2 and (\omega = 314) rad/s gives (\varepsilon_0 = 200 \times 0.4 \times 0.05 \times 314 = 4 \times 314 = 1256) V. The option 628 V halves the result by using (\pi\times50) instead of (2\pi\times50). The option 400 V drops the angular-speed factor and keeps only (NBA) scaled. The option 200 V uses the turns count alone. This is the NCERT generator relation in which peak EMF scales with rotation rate, arising because the flux varies as (\Phi = NBA\cos\omega t) and its time derivative introduces the factor (\omega). The instantaneous EMF is therefore (\varepsilon = NBA\omega\sin\omega t), a pure sinusoid whose maximum is the peak value computed above and whose RMS value would be smaller by a factor of (\sqrt{2}). A plausibility check confirms the units (T·m^2·rad/s per turn, times turns) give volts, and a high peak is entirely reasonable for a many-turn coil spinning rapidly in a moderate field, which is why practical generators use many turns and high rotation speeds.
This medium difficulty physics question is from the chapter electromagnetic induction and alternating currents, covering the topic of flux linkage in a rotating coil. It appeared in the 2025 exam.
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