Physics · Ch 3 — Magnetism and Magnetic Effects of Electric Current
Magnetic Field Due to a Long Current Carrying Solenoid
Magnetic Field Due to a Long Current Carrying Solenoid
To find the field inside a long solenoid of turns over length (turns per unit length ), choose a rectangular Amperian loop with side (length , taken equal to the full solenoid length for convenience) lying inside the solenoid parallel to the axis, side lying entirely outside, and sides , crossing perpendicular to the field. Since outside, the integral along is zero; since and are perpendicular to , those two integrals are also zero (); only the leg survives, contributing . Ampere's law with the enclosed current (all turns pass through the loop) then gives , so
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What this figure shows. A long solenoid is drawn in cross-section with its field lines pointing along its axis inside and spreading out negligibly outside. A rectangular Amperian loop abcd is superimposed so that side ab (length h) lies inside the solenoid parallel to the axis, side cd lies entirely outside, and sides bc and da cross from inside to outside, perpendicular to the field -- the specific loop shape used to isolate just …
Worked out. Starting from B = mu0 N I / L for a solenoid of N turns over length L: (a) if only the length doubles (L to 2L) with N fixed, the new field is B' = mu0 N I/(2L) = B/2, i.e. it halves. (b) If both L and N double together, the turns-per-length ratio N/L is unchanged, so B' = mu0 (2N) I/(2L) = mu0 N I/L = B, i.e. it stays the same. (c) If N doubles but L stays fixed, B' = mu0(2N)I/L = 2B, i.e. it doubles. Comparing the three, B(L,2N) > B(L,N) > B(2L,N): packing more turns into the same length increases the field, stretching the same turns over a longer length decreases it, and sca …