Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

A coil is formed by winding 250turns of insulated 16-gauge copper wire (diameter = 1.3 mm) in a single layer on a cylindrical form of radius 12cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Use Table 26-1.)

The resistance of the coil is 2.4 $Ω$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Number of turns of coil,n = 250
Diameter of the wire, $d = 1.3 mm or 1.3 \times 10^{- 3} m$
Radius of coil, $R = 12 cm or 12 \times 10^{- 2} m$

Step 2: Understanding the concept of the resistance

The voltage is directly proportional to the current and the constant of proportionality is called Resistance. This is called Ohm’s law.

We can find the total length of the coil by using the radius of the coil. We can also find the area of the cross-section of the wire from its diameter. Then, by using these values in the formula for resistance, we can find the resistance of the copper coil.

Formulae:

The resistance of a wire due, $R = \frac{p L}{A}$ …(i)

Here, R is resistance, p is resistivity, L is length, A is the area of cross-section.

The circumference of the circle, $L = 2 π R$ …(ii)

is circumference, is the radius of the circle.

The cross-sectional area of a circle, $A = {πR}^{2}$ …(iii)

Step 3: Calculation of the resistance of the coil

Length of one turn of the coil can be given using the circumference of the circle.

So, the length of 250 turns of the coil using equation (ii) would be,

$L = 250 \times (2 πR) = 250 \times 2 (3.142) (12 \times 10^{- 2} m) = 188.49 m$

Similarly, the cross sectional area of the wire is given using equation (iii) as follows:

$A = π \frac{d^{2}}{4} (∴ r = d / 2) = (3.142) \frac{{(1.3 \times 10^{- 3} m)}^{2}}{4} = 1.327 \times 10^{- 6} m^{2}$

Resistivity of copper is $p = 1.69 \times 10^{- 8} Ωm$

Now, using these above values in equation (i); the resistance of the copper coil can be calculated as follows:

$R = \frac{(1.69 \times 10^{- 8} Ωm) (188.49 m)}{1.327 \times 10^{- 6} m^{2}} = 2.4 Ω$

Therefore, the resistance of the copper coil is $2.4 Ω$ .

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

A coil is formed by winding 250turns of insulated 16-gauge copper wire (diameter = 1.3 mm) in a single layer on a cylindrical form of radius 12cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Use Table 26-1.)

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of the resistance

Step 3: Calculation of the resistance of the coil

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

A coil is formed by winding 250turns of insulated 16-gauge copper wire (diameter = 1.3 mm) in a single layer on a cylindrical form of radius 12cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Use Table 26-1.)

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of the resistance

Step 3: Calculation of the resistance of the coil

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics