Two ideal solenoids of radii R and 4 R , respectively, have n turns per meter, and each carries a current of I . Both currents flow in the same direction. The small‑radius solenoid is placed inside the large-radius solenoid so that their axes of symmetry are parallel and separated by a distance d = 2 R . Neglecting any magnetic screening effects, express the magnetic field strength on the axis of the small-radius solenoid analytically in terms of the quantities given and the magnetic permeability of a vacuum, μ 0 .

The magnetic field inside de solenoid is constant. In the case of a small-radius solenoid inside a large-radius solenoid, the magnetic field inside the small-radius solenoid is the magnetic field generated by itself plus the magnetic field generated by the large-radius solenoid. (The radius of the solenoids does not have to be with the instensity of the magnetic field):

BT = Bs + Bl

Bs: magnetic fiel of the small-radius solenoid

Bl: magnetic fiel of the large-radius solenoid

Hence:

BT = 2*μo*n*I

topeadeniran2 topeadeniran2 · Answer 2 · 2022-03-29T18:23:41+02:00

The magnetic field strength based on the information is 2UonI.

How to calculate the magnetic field strength?

It should be noted that the assumptions made in this scenario will be that the magnetic field just outside the solenoid is zero.

Also, the magnetic field inside the solenoid is uniform. Therefore, the magnetic field strength will be:

= UonI + UonI

= 2UonI

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