CCommons
Search✍️ WriteSign in
⚛️ NEET Physics❓ Question

AC generator doubts: what changes if speed doubles, and where does the electrical energy actually come from?

Nisha Verma12th Sample

17 Sept 2026 · 1 min read

Boards + NEET so I need this clean. NCERT gives e=NBAωsin⁡ωte = NBA\omega\sin\omega t for a coil of N turns, area A, rotating at angular speed ω in a field B.

I did a numerical: coil 50 turns, area 0.02 m², B = 0.4 T, rotating at 3000 rpm, find peak emf. First I put ω = 3000 and got a crazy answer. Then I realised rpm needs converting: 3000 rpm = 50 rev/s, so ω=2π×50≈314\omega = 2\pi\times50 \approx 314 rad/s and

e0=50×0.4×0.02×314≈126 Ve_0 = 50\times0.4\times0.02\times314 \approx 126\ \text{V}

OK that part is fixed. But I still have:

(a) If the speed of rotation is doubled, what happens to the peak emf and to the frequency of the AC?

(b) My friend says emf is maximum when the coil is perpendicular to the field because "maximum lines are passing". My notes say emf is zero at that position. Who is right?

(c) Current flows in the coil so there must be a magnetic force on it. Does it oppose the rotation? Then where is the energy coming from? Feels like free electricity.

WhatsAppPost on X

7 answers & replies

Sign in to join the conversation.

  • [+]Harsh Gupta12thSample· 1mo ago

    (a) e0=NBAωe_0 = NBA\omega so peak emf doubles. Frequency is f=ω/2πf = \omega/2\pi so that doubles as well. Both go with ω. Your 50 Hz becomes 100 Hz and the bulb in a power cut inverter says "bhai mujhe mat ghuma" 😂

    Also the number of turns only changes the peak, never the frequency. If an MCQ asks "double N", frequency stays same.

  • [+]Ishita GoyalDropperSample· 1mo ago

    (b) Your notes are right, friend is wrong. When the coil is perpendicular to B, the normal is parallel to B so flux is maximum, but the two long sides are moving along the field direction, so they don't cut any field lines and emf is zero.

    When the plane of the coil is parallel to B, flux is zero but the sides cut the field at 90° so emf is max. I keep a mini table: normal ∥ B → Φ max, e = 0. Normal ⊥ B → Φ = 0, e max.

  • [+]Aman YadavDropperSample· 1mo ago

    (c) Yes. Current in the coil in a magnetic field gives a torque that opposes rotation. That is Lenz's law at work. So you have to keep applying a mechanical torque to rotate the coil. The energy comes from whatever turns the coil: water, steam, wind, your hand.

    No free electricity. Mechanical energy in, electrical out. More current drawn, harder to rotate.

    • [+]Nisha Verma12thSample· 1mo agoAuthor

      okay that makes sense. so more load current = harder to turn. that also explains why a cycle dynamo light feels heavier to pedal when it's on lol

  • [+]Pooja Iyer12thSample· 1mo ago

    Small add: the direction of induced current in the generator is found by Fleming's right hand rule (thumb = motion of the conductor, forefinger = field, middle finger = current). Mnemonic I use: "right hand = generator, left hand = motor". Left hand for motor because force is what you get.

    And for frequency with a simple two-pole alternator, f=rpm/60f = \text{rpm}/60. So 3000 rpm gives exactly 50 Hz, which is why your numerical used such a neat number.

  • [+]Dev Agarwal12thSample· 1mo ago

    For (a) I think only the peak emf doubles. The frequency stays 50 Hz because it is fixed by the grid, no? Power station frequency is always 50.

    • [+]Ishita GoyalDropperSample· 1mo ago

      Dev, that is the grid's choice, not physics. The grid keeps the generators at a fixed speed on purpose so everybody gets 50 Hz. If you spin one coil faster in a lab, frequency goes up right away, f=ω/2πf = \omega/2\pi. Harsh's answer is the correct one for the question.

More from Nisha Verma

Keep reading