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Q.Obtain an expression for the magnetic field on the axis of a current carrying very long solenoid using Ampere's circuital law. Draw the necessary diagram.

Rajasthan RbseRajasthan Board Senior Secondary Examination 2023Subjective· 3mImportance★★★★★
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Figure — Stem asks for the diagram of a long solenoid with the Amperian loop used to find its axial field; the catalog
Figure — Stem asks for the diagram of a long solenoid with the Amperian loop used to find its axial field; the catalog

Applying Ampere's circuital law to a rectangular Amperian loop, half inside and half outside a long solenoid, gives the uniform axial field B = mu_0 n I.

Setup: consider a long solenoid of nn turns per unit length carrying a steady current II. For an ideal long solenoid, the magnetic field is essentially uniform and parallel to the axis inside the solenoid, and essentially zero just outside it.

Amperian loop: choose a rectangular loop abcdabcd such that side abab (length LL) lies inside the solenoid parallel to the axis, side cdcd lies outside the solenoid (also parallel to the axis, where B≈0B\approx0), and the two sides bcbc, dada are perpendicular to the axis, partly inside and partly outside.

Applying Ampere's law, ∮B⃗⋅dl⃗=μ0Ienc\oint \vec B\cdot d\vec l = \mu_0 I_{enc}, around loop abcdabcd:

  • Along abab (inside, parallel to B⃗\vec B): contributes BLB L.
  • Along cdcd (outside, where B≈0B \approx 0): contributes 00. …

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