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Q.Write Ampere's circuital law, with the help of this law establish an expression for the intensity of the magnetic field inside the solenoid with the number of turns and flowing current. [1+4=5] OR Describe the moving coil galvanometer under the following points:

(i) Labelled diagram
(ii) Principle [2+3=5]
Chhattisgarh CgbseCGBSE Intermediate Board 2026Subjective· 5mImportance★★★★★
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Ampere's law relates the closed-loop line integral of BB to the enclosed current; choosing a rectangular Amperian loop straddling the solenoid's winding gives B=μ0nIB=\mu_0 nI inside a long solenoid.

Ampere's circuital law: The line integral of the magnetic field B⃗\vec B around any closed loop equals μ0\mu_0 times the total (conduction) current enclosed by that loop:

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

Field inside a solenoid: Consider a long solenoid with nn turns per unit length carrying a steady current II. For an ideal (very long, tightly wound) solenoid, the field is essentially uniform and parallel to the axis inside the solenoid, and essentially zero outside it.

Choose a rectangular Amperian loop PQRSPQRS: side PQPQ (length ll) lies inside the solenoid, parallel to the axis; side RSRS lies far outside the solenoid; the sides QRQR and SPSP are perpendicular to the axis, partly inside and partly outside.

∮B⃗⋅dl⃗=∫PQB⃗⋅dl⃗+∫QRB⃗⋅dl⃗+∫RSB⃗⋅dl⃗+∫SPB⃗⋅dl⃗\oint\vec B\cdot d\vec l=\int_{PQ}\vec B\cdot d\vec l+\int_{QR}\vec B\cdot d\vec l+\int_{RS}\vec B\cdot d\vec l+\int_{SP}\vec B\cdot d\vec l

  • Along PQPQ (inside, parallel to BB): contributes BlBl.
  • Along RSRS (outside, where B≈0B\approx0): contributes 00. …

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