A coil is formed by winding 330 turns of insulated 16 gauge copper wire (diameter = 1.0 mm) in a single layer on a cylindrical form of radius 17 cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Take the resistivity of copper to be 1.69 x 10-8 ohm-m.)
A coil is formed by winding 330 turns of insulated 16 gauge copper wire (diameter = 1.0 mm) in a single layer on a cylindrical form of radius 17 cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Take the resistivity of copper to be 1.69 x 10-8 ohm-m.)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
Please type the answer instead of hand writting thanks
![**Problem Statement:**
A coil is formed by winding 330 turns of insulated 16-gauge copper wire (diameter = 1.0 mm) in a single layer on a cylindrical form of radius 17 cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Take the resistivity of copper to be \(1.69 \times 10^{-8}\) ohm-m.)
---
**Solution:**
To calculate the resistance of the coil, use the formula for the resistance of a conductor:
\[ R = \rho \frac{L}{A} \]
where:
- \( R \) is the resistance (ohms)
- \( \rho \) is the resistivity of the material (ohm-meters), given as \(1.69 \times 10^{-8}\) ohm-m
- \( L \) is the length of the wire (meters)
- \( A \) is the cross-sectional area of the wire (square meters)
### Step-by-Step Calculation:
1. **Determine the length of the wire:**
Each turn of the wire wraps around the cylindrical form. The circumference of one turn is:
\[ \text{Circumference} = 2 \pi r \]
where \( r \) is the radius of the cylindrical form (0.17 meters).
\[ \text{Circumference} = 2 \pi \times 0.17 = 1.068 \text{ meters} \]
Since there are 330 turns:
\[ L = 330 \times 1.068 = 352.44 \text{ meters} \]
2. **Calculate the cross-sectional area \(A\):**
The diameter of the wire is 1.0 mm, so the radius \( r_w \) is 0.5 mm (0.0005 meters).
\[ A = \pi r_w^2 = \pi \times (0.0005)^2 = 7.85 \times 10^{-7} \text{ square meters} \]
3. **Calculate the resistance \(R\):**
\[ R = \rho \frac{L}{A} = (1.69 \times 10^{-8}) \frac{352.44}{7.85 \times 10^{-7}} \]
\[ R \approx 0.0076 \text{ ohms} \]
---
**Results:**
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe462800b-ac16-4590-bafb-82d68c005ec5%2F791acf3c-d3ef-4c2f-b76c-186a1b7ae95f%2F84gr5bk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A coil is formed by winding 330 turns of insulated 16-gauge copper wire (diameter = 1.0 mm) in a single layer on a cylindrical form of radius 17 cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Take the resistivity of copper to be \(1.69 \times 10^{-8}\) ohm-m.)
---
**Solution:**
To calculate the resistance of the coil, use the formula for the resistance of a conductor:
\[ R = \rho \frac{L}{A} \]
where:
- \( R \) is the resistance (ohms)
- \( \rho \) is the resistivity of the material (ohm-meters), given as \(1.69 \times 10^{-8}\) ohm-m
- \( L \) is the length of the wire (meters)
- \( A \) is the cross-sectional area of the wire (square meters)
### Step-by-Step Calculation:
1. **Determine the length of the wire:**
Each turn of the wire wraps around the cylindrical form. The circumference of one turn is:
\[ \text{Circumference} = 2 \pi r \]
where \( r \) is the radius of the cylindrical form (0.17 meters).
\[ \text{Circumference} = 2 \pi \times 0.17 = 1.068 \text{ meters} \]
Since there are 330 turns:
\[ L = 330 \times 1.068 = 352.44 \text{ meters} \]
2. **Calculate the cross-sectional area \(A\):**
The diameter of the wire is 1.0 mm, so the radius \( r_w \) is 0.5 mm (0.0005 meters).
\[ A = \pi r_w^2 = \pi \times (0.0005)^2 = 7.85 \times 10^{-7} \text{ square meters} \]
3. **Calculate the resistance \(R\):**
\[ R = \rho \frac{L}{A} = (1.69 \times 10^{-8}) \frac{352.44}{7.85 \times 10^{-7}} \]
\[ R \approx 0.0076 \text{ ohms} \]
---
**Results:**
-
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON