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Derive the expression for the self induc...

Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.

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To derive the expression for the self-inductance \( L \) of a long solenoid with cross-sectional area \( A \), length \( l \), and having \( n \) turns per unit length, we can follow these steps: ### Step 1: Define the parameters of the solenoid Let: - \( R \) = radius of the solenoid - \( l \) = length of the solenoid - \( A \) = cross-sectional area of the solenoid = \( \pi R^2 \) - \( n \) = number of turns per unit length ...
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Explore conceptually related problems

(a) Define self inductance. Write its S.I. units (b) Derive and expression for self inductance of a long solenoid of length l cross-sectional area A having N number of turns.

The self inductance of a solenoid of length L, area of cross-section A and having N turns is-

Knowledge Check

  • The self inductance of a solenoid that has a cross-sectional area of 1 cm^(2) , a length of 10 cm and 1000 turns of wire is

    A
    `0.86 mH`
    B
    `1.06 mH`
    C
    `1.26 mH`
    D
    `1.46 mH`
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    Derive the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid carrying a steady current 1. How does this magnetic energy per unit volume compare with the electrostatic energy density stored in a parallel plate capacitor ?

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