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NCERT Exemplar · Q17

Q.There are two coils AA and BB separated by some distance. If a current of 2 A2\ \text{A} flows through AA, a magnetic flux of 10−2 Wb10^{-2}\ \text{Wb} passes through BB (no current through BB). If no current passes through AA and a current of 1 A1\ \text{A} passes through BB, what is the flux through AA?

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Mutual inductance is symmetric: the flux through coil A when 1 A flows in coil B equals the flux through coil B when 2 A flows in coil A, scaled by the current ratio. The answer is 5×10−3 Wb5 \times 10^{-3}\ \text{Wb}.

The key idea here is mutual inductance — a property of two coils that links the current in one to the flux through the other. When current changes in coil A, it induces a flux in coil B, and vice versa. The constant that relates them, MM, is the same in both directions. That symmetry is the entire backbone of this problem.

Let’s work through it step by step.

  1. Define mutual inductance.

    For two coils AA and BB, the mutual inductance MM is defined by:

    • When current IAI_A flows in AA, the flux through BB is ΦB=MIA\Phi_B = M I_A.
    • When current IBI_B flows in BB, the flux through AA is ΦA=MIB\Phi_A = M I_B.

    The same MM appears in both equations — this is a fundamental result from electromagnetic theory (and follows from energy conservation / reciprocity).

  2. Use the first situation to find MM.

    We are told: IA=2 AI_A = 2\ \text{A}, and the flux through BB is ΦB=10−2 Wb\Phi_B = 10^{-2}\ \text{Wb}.

    From ΦB=MIA\Phi_B = M I_A:

M=ΦBIA=10−22=5×10−3 HM = \frac{\Phi_B}{I_A} = \frac{10^{-2}}{2} = 5 \times 10^{-3}\ \text{H}

(henry, the unit of inductance).

  1. Now apply the second situation. Here, IB=1 AI_B = 1\ \text{A} and we want ΦA\Phi_A. Using the same MM: …

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