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A 36.5 mA current is carried by a uniformly wound air-core solenoid with 430 turns, a 18.5 mm diameter, and 11.5 cm length. (a) Compute the magnetic field inside the solenoid. (b) Compute the magnetic flux through each turn. Tm2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current?i) magnetic field inside the solenoid ii) magnetic flux through each turn inductance of the solenoid

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Answer:

Step-by-step explanation:

a )

magnetic field inside the solenoid B = μ₀ n I where n is no of turns per unit length , I is current and μ₀ = 4 π x 10⁻⁷ .

Putting the values in the equation

B = 4 π x 10⁻⁷ x (430 / .115 ) x 36.5 x 10⁻³

= 1.7 x 10⁻⁴ T .

b ) magnetic flux through each turn

= B x A where A is cross sectional area of solenoid .

= 1.7 x 10⁻⁴ x π x 9.25² x 10⁻⁶

= 456.73 x 10⁻¹⁰ Tm² .

c ) Inductance of solenoid

L = flux associated with all turns / current

= 456.73 x 10⁻¹⁰ x 430 / (36.5 x 10⁻³)

= 5381 x 10⁻⁷

= 538 x 10⁻⁶ H

= 538 μH .

d )

magnetic field inside the solenoid depends upon current

magnetic flux through each turn depends upon current

inductance of the solenoid does not depend upon current because current is divided from total flux with solenoid.

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