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Darivation of solenoid

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Darivation of solenoid

    • Posted by Aman Kumar 7 years, 1 month ago

    CBSE > Class 12 > Physics
    • 1 answers

    Sanskar Sharma 7 years, 1 month ago

    Solenoid Consider a rectangular Amperean loop PQRS. Along PQ the field is zero. Along transverse sections PS and QR, the field component is zero. Let the magnetic field along SR be B. Loop ∮ Bdl cosθ = μ0i ∮ B→dl→ = ∮ Bdl cosθ = R∫S Bdl cos0 + Q∫R Bdl cos90o + P∫Q Bdl cos180 + S∫P Bdl cos90o = B R∫S dl + 0 + 0 + 0 = Bl Current passing through loop – ??? Thus, the number of turns per unit length is ‘n’, then the total number of turns is ‘nh’. The enclosed current is Ie = I(?ℎ) where I is the current in the solenoid. From Ampere’s circuit law BL = μ0Ie, Bh = μ0I(nh) B = μ0nI

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