Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 mm 9.49 mm and its length is 2.09 cm 2.09 cm . What is the inductance, in microhenrys, of Tarik's solenoid?

143μH

The inductance (L) of a coil wire (e.g solenoid) is given by;

L = μ₀N²A / l                 (i)

Where;

l = the length of the solenoid

A = cross-sectional area of the solenoid

N= number of turns of the solenoid

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

From the question;

N = 183 turns

l = 2.09cm = 0.0209m

diameter, d = 9.49mm = 0.00949m

But;

A = π d² / 4                     [Take π = 3.142 and substitute d = 0.00949m]

A = 3.142 x 0.00949² / 4

A = 7.1 x 10⁻⁵m²

Substitute these values into equation (i) as follows;

L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209           [Take π = 3.142]

L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209

L = 143 x 10⁻⁶ H

L = 143 μH

Therefore the inductance in microhenrys of the Tarik's solenoid is 143

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