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Obtain an expression for the self-inductance of a solenoid. - Physics

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प्रश्न

Obtain an expression for the self-inductance of a solenoid.

व्युत्पत्ति
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उत्तर

  1. Consider a current I established in the windings (turns) of a long solenoid. The current produces a magnetic flux `phi_"B"` through the central region.
  2. The inductance of the solenoid is given by,
    L = `("N"phi_"B")/"I",`
    where N = the number of turns,
    ΦB = magnetic flux linkage. 
  3. The flux linkage for a length l near the middle of the solenoid is,
    B = (nl) `(vec"B". vec"A")` = nlBA, (for θ = 0°),
    where n = the number of turns per unit length,
    B = magnetic field
    A = the cross-sectional area of the solenoid.
  4. The magnetic field inside the solenoid is given as, B = `mu_0"ni"`
  5. Hence, L = `("N"phi_"B")/"i"`
    = `(("nl")"BA")/"i"`
    = `("nl"(mu_0"ni")"A")/"i"`
    ∴ L = μ0n2lA 
    where, Al is the interior volume of solenoid. 
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अध्याय 12: Electromagnetic Induction - Short Answer II

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