A solenoid 2.5 cm in diameter and 30 cm in length has 4800 turns and carries a current of 2.0 A. Part A Calculate the magnetic flux through the circular cross-sectional area of the solenoid. Hint: Assume this is a very long solenoid and use the simplified magnetic field formula for an infinite solenoid.

Ø = 0.034Wb

Explanation:

A solenoid is idealized as a certain number of current loops of diameter d in series.

• The current loops create a magnetic field. This field permeates each loop, ie, there is magnetic flux through each loop.

Determine the number of turns from the length and given winding rate.

• Calculate the strength of the magnetic field produced by the solenoid, then the flux through one turn (winding).

Please find the attached file for the solution.

it takes about 30 days for it to break down