Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

An AC appliance cord has its hot and neutral wires separated by $3.00 mm$ and carries a $5.00 - A$ current. (a) What is the average force per meter between the wires in the cord? (b) What is the maximum force per meter between the wires? (c) Are the forces attractive or repulsive? (d) Do appliance cords need any special design features to compensate for these forces?

(a) The average force per meter is calculated as $1.67 \times 10^{3} N / m$ .

(b) Maximum force per meter is calculated as $3.34 x 10^{- 3} N / m$ .

(d) No special design features are needed for appliance cords.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Current in the wires $I_{1} and I_{2} is 5.00 A$

The separation between the wires is $r = 3.00 mm (\frac{1 m}{1000 mm}) = 3.00 \times 10^{- 3} m$

Step 2: Definition of force

In physics, a force is an effect that can alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. Intuitively, the force can be described as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Step 3: Calculate the current in the wires

(a)

To determine the current in the wires, we will use the formula for force per unit length that can be expressed as,

$\frac{F}{L} = \frac{μ {}_{0}l {}_{1}l_{2}}{2 πr}$ ……………(1)

Where $l_{1} and l_{2}$ is the current in the wires, $r$ is the separation between the wires, and $μ_{0}$ is the permeability of free space.

The current is the same in both the wires as it is one cord $l_{1} = l_{2} = l$ .

$\frac{F}{L} = \frac{4 π \times 10^{- 7} N / A^{2} \times {(5.00 A)}^{2}}{2 π \times (3 \times 10^{- 3} m)} = 1.67 \times 10^{- 3} N / m$

Therefore, the average force per meter is $1.67 \times 10^{- 3} N / m$ .

Step 4: Evaluating the maximum force per meter

(b)

The first wire exerts a force on the second wire and vice versa. As a result, the greatest force on the wires is equal to the sum of the two forces.

$F_{m a x} = 2 \times (1.67 \times 10^{- 3} N / m) = 3.34 \times 10^{- 3} N / m$

Therefore, the maximum force per meter is $3.34 \times 10^{- 3} N / m$

Step 5: Forces attractive or repulsive

(c)

Since the currents in the two wires are flowing in opposite directions, the force between them will be repellent.

Step 6: Any special design feature needed

(d)

Since the forces between the wires are so minimal, no special design elements are required to compensate for them.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Definition of force

Step 3: Calculate the current in the wires

Step 4: Evaluating the maximum force per meter

Step 5: Forces attractive or repulsive

Step 6: Any special design feature needed

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Definition of force

Step 3: Calculate the current in the wires

Step 4: Evaluating the maximum force per meter

Step 5: Forces attractive or repulsive

Step 6: Any special design feature needed

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