A solenoid of length 2.9 cm and diameter 0.71 cm is wound with 165 turns per cm.

Question

A solenoid of length 2.9 cm and diameter 0.71 cm is wound with 165 turns per cm. When the current through the solenoid is 0.22 A, what is the magnetic flux through one of the windings of the solenoid?

Explanation / Answer

Assume the solenoid is air-cored, so permeability mu0 = 4pi.e-7 H/m Also treat the solenoid as magnetically "long" since length/diameter > 3 (This means we ignore "end effects" and assume the flux is constant throughout the core) So magnetic field strength B = 4pi.e-7 x 16500 turns per m x 0.22A = 4.56e-3 Tesla Cross-sectional area of the solenoid = pi.(0.00355m)^2 = 3.95e-5 m^2 So magnetic flux (phi) = 4.56e-3 T x 3.95e-5 m^2 = 1.8e-7 Weber This is the flux through one of the windings, and also the flux through every winding.

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