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Q.Write Ampere's circuital law and by using this law derive the expression for intensity of magnetic field inside a current-carrying solenoid. OR Explain moving coil galvanometer on the following points:

(i) Labelled diagram
(ii) Principle
Chhattisgarh CgbseCGBSE Intermediate Board 2022Subjective· 5mImportance★★★★★
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Ampere's law equates the line integral of B around a closed loop to μ₀ times the enclosed current; using a rectangular loop partly inside and partly outside a long solenoid gives the uniform interior field B = μ₀nI.

Ampere's circuital law: The line integral of the magnetic field B⃗\vec B around any closed loop is equal to μ₀ times the total current enclosed by that loop:

∮B⃗⋅dl⃗=μ0Ienc\oint \vec B \cdot d\vec l = \mu_0 I_{enc}

Field inside a long current-carrying solenoid:

Consider a long solenoid with n turns per unit length, carrying a steady current I. For an ideal long solenoid, the magnetic field inside is strong, uniform, and directed along the axis, while the field just outside the solenoid is negligibly small (≈ 0).

Choose a rectangular Amperian loop PQRS such that:

  • Side PQ (length L) lies inside the solenoid, parallel to the axis, where the field is uniform and equal to B.
  • Side RS lies outside the solenoid, where the field is ≈ 0.
  • Sides QR and SP are perpendicular to the axis (partly inside, partly outside); along these, the magnetic field (wherever nonzero, i.e. inside) is perpendicular to dl⃗d\vec l, so B⃗⋅dl⃗=0\vec B \cdot d\vec l = 0 on these sides, and outside the field is ≈ 0 anyway. …

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