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⚛️ NEET Physics❓ Question

Flux = BA cosθ, but which angle is θ? And why is emf zero when flux is maximum?

Tanvi Deshmukh12th Sample

9 Oct 2026 · 1 min read

ok so I made a really neat table in my notes for Faraday's law but the very first line is breaking my brain. NCERT defines flux as ΦB=B⋅A=BAcos⁡θ\Phi_B = \mathbf{B}\cdot\mathbf{A} = BA\cos\theta.

In the example I solved, a coil is placed with its plane making 30° with a uniform field of 0.5 T, area 0.02 m². I took θ=30°\theta = 30° and got

Φ=0.5×0.02×cos⁡30°≈8.7×10−3 Wb\Phi = 0.5 \times 0.02 \times \cos 30° \approx 8.7\times 10^{-3}\ \text{Wb}

The answer key says 5×10−35\times10^{-3} Wb. 😭

Second thing. In the rotating coil figure the flux is maximum at one instant, and in my notes I wrote "flux max → emf max, more flux = more induced emf". But the emf graph clearly shows zero emf at that instant. I copied both graphs in different colours and they just don't match what I wrote.

So two doubts:

  1. which angle is θ, exactly?
  2. emf is the rate of change of flux, fine, but why does zero emf come with maximum flux?

I want one clean way to remember this so I don't lose marks in boards and NEET both.

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7 answers & replies

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  • [+]Siddharth RaoDropperSample· 10d ago

    θ is the angle between B and the area vector (the normal to the coil), not the plane. Your plane is at 30° to the field so the normal is at 60°. Φ=0.5×0.02×cos⁡60°=5×10−3\Phi = 0.5\times0.02\times\cos60° = 5\times10^{-3} Wb, matches the key.

    I lost 4 marks in a test last year on exactly this. Now the first thing I do in any flux question is draw the normal arrow on the coil and mark the angle from it. If the problem says "plane parallel to field" the flux is zero, because the normal is at 90°.

  • [+]Vivek Mishra12thSample· 10d ago

    For the second doubt I think emf is more when flux is more because the coil is catching more field lines at that time. That is why the emf graph has a peak in the middle?

  • [+]Neel Chatterjee12thSample· 10d ago

    Faraday's law is ε=−N dΦdt\varepsilon = -N\,\dfrac{d\Phi}{dt}. It cares about how fast flux changes, not how big it is.

    For a coil rotating at constant ω with normal along B at t = 0: Φ=BAcos⁡ωt\Phi = BA\cos\omega t. So

    ε=−NdΦdt=NBAωsin⁡ωt\varepsilon = -N\frac{d\Phi}{dt} = NBA\omega\sin\omega t

    Cos is at its peak when sin is zero. Flux max means the graph of Φ against t is flat at the top for a moment, slope zero, so emf zero. Flux is zero when the Φ curve is steepest, so emf is max there.

    Easy picture: height of a ball thrown up vs its speed. At the top the height is max but the velocity is zero.

    • [+]Tanvi Deshmukh12thSample· 10d agoAuthor

      the ball analogy finally made it click, thank you!! I'm writing "emf = slope of Φ–t graph" in red on top of my table 😭

    • [+]Vivek Mishra12thSample· 9d ago

      Okay I understand now. My answer was wrong, sorry. Emf is slope not height.

  • [+]Kavya NaikDropperSample· 10d ago

    Tanvi, a tiny trick that saved me: whenever you see the words "plane of the coil" the angle given is with the plane, so convert. Use the cos with the normal or sin with the plane:

    Φ=BAsin⁡α\Phi = BA\sin\alpha, where α is the angle between B and the plane.

    Then 30° with plane gives 0.5×0.02×sin⁡30°=5×10−30.5\times0.02\times\sin30° = 5\times10^{-3} Wb directly. Both are same thing, just pick one and stick to it, don't mix. amma used to say "mix two recipes and the sambar is ruined" 😂

  • [+]Rahul DubeyDropperSample· 10d ago

    Two small things to add for NEET. First, unit of flux is weber, 1 Wb=1 T m21\ \text{Wb} = 1\ \text{T m}^2. Second, for a coil of N turns the flux linkage is NΦN\Phi, so the emf has the N, but "flux through the coil" in the question usually means one turn unless they say otherwise.

    Also the minus sign is Lenz's law: induced current opposes the change in flux, not the flux itself. That one idea explains why emf is zero at max flux: the flux isn't changing at that instant, so there is nothing for the induced current to oppose.

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