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.

Question

Alonso Singh

Physics

17 February, 09:33

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.

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Reagan Wallace · Answer 1

The formula for the calculation of the magnetic field inside a solenoid is

B = μo*n*I

where

μo: vacuum permeability

n: turns per meter

I: current

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