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1. A coil is formed by winding 250 turns of insulated 16-gauge copper wire, that has a diameter d = 1.3 mm, in a single layer on a cylindrical form of radius 12 cm. What is the resistance of the coil? Neglect the thickness of the insulation and the resistivity of copper is ???? = 1.69 × 10−8 Ω ∙ m.

User Florida
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Final answer:

Since the coil is formed by winding 250 turns, the length of the wire will be equal to the circumference of the cylindrical form multiplied by the number of turns: The resistance of the coil is approximately 0.191 Ω.

Step-by-step explanation:

To calculate the resistance of a coil, we need to use the formula:

R = (ρ * L) / A

Where:

R is the resistance of the coil

ρ is the resistivity of copper

L is the length of the wire

A is the cross-sectional area of the wire

First, we need to find the length of the wire.

Since the coil is formed by winding 250 turns, the length of the wire will be equal to the circumference of the cylindrical form multiplied by the number of turns:

L = 2πr * N

Where:

r is the radius of the cylindrical form

N is the number of turns

Substituting the values into the formula, we get:

L = 2π * 0.12 m * 250 = 150 m

Next, we need to calculate the cross-sectional area of the wire.

The diameter of the wire is given, so we can use the following formula to find the area:

A = πr^2

Substituting the values into the formula, we get:

A = π * (0.0013 m / 2)^2 = 1.33 × 10^-6 m^2

Now, we can calculate the resistance by substituting the values into the resistance formula:

R = (1.69 × 10^-8 Ω ∙ m * 150 m) / (1.33 × 10^-6 m^2)

= 0.191 Ω

Therefore, the resistance of the coil is approximately 0.191 Ω.

User Birol Efe
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