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An air-filled 39.1 μH solenoid has a length 4.0 cm and a cross-sectional area 0.60 cm2. How many turns are in this solenoid?
a.21000
b.144
c.12
d.120
e.1200

Answer :

No. of turns in the solenoid is an option (b) 144.

The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.

Self-inductance of solenoid = 39.1 μH

                                           = 39.1 × [tex]10^-^6[/tex] H

Length of the solenoid = 4.0 cm

Cross-sectional area = 0.60 cm²

Expression for the self-inductance of a coil ;

L = µ₀N²A / [tex]l[/tex]

where,

L = Self- Inductance

N = No. of turns.

A = Cross-sectional area

[tex]l=[/tex] Length of the solenoid

L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0

39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²

N² = 2.07 × [tex]10^6[/tex]

N = 144

Therefore, the no. of turns of the solenoid is 144.

Learn more about self-inductance here:

https://brainly.com/question/25484149

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