Skip to content
Question 94 of 132

Q.An a.c. generator consists of a coil of 10,000 turns and of area 100 cm2^2. The coil rotates at an angular speed of 140 rpm in a uniform magnetic field of 3.6×10−23.6\times10^{-2} T. Find the maximum value of the emf induced.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2017Subjective· 3mImportance★★★★★
71% · 94/132 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The maximum induced emf of this AC generator works out to about 52.8 V52.8\ V.

For an AC generator with a coil of NN turns, area AA, rotating with angular speed ω\omega in a uniform magnetic field BB, the induced emf is e=NBAωsin⁡ωte = NBA\omega\sin\omega t, so the maximum (peak) emf is

E0=NBAωE_0 = NBA\omega

Given: N=10,000N = 10{,}000 turns; A=100 cm2=100×10−4 m2=1×10−2 m2A = 100\ cm^2 = 100\times10^{-4}\ m^2 = 1\times10^{-2}\ m^2; B=3.6×10−2 TB = 3.6\times10^{-2}\ T; rotational speed =140= 140 rpm.

Step 1 — convert rpm to angular speed in rad/s:

ω=2π×14060=879.6560≈14.66 rad/s\omega = \frac{2\pi \times 140}{60} = \frac{879.65}{60} \approx 14.66\ rad/s

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.