 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q54PE

Expert-verified Found in: Page 813 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # An AC appliance cord has its hot and neutral wires separated by ${\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{mm}}$ and carries a ${\mathbf{5}}{\mathbf{.}}{\mathbf{00}}{\mathbf{-}}{\mathbf{A}}$ 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 $1.67×{10}^{3}\mathrm{N}/m$.

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

(c) The force will be repulsive.

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

See the step by step solution

## Step 1: Given data

Current in the wires ${\mathrm{I}}_{1}\mathrm{and}{\mathrm{I}}_{2}\mathrm{is}5.00\mathrm{A}$

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

## 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,

$\frac{\mathrm{F}}{\mathrm{L}}=\frac{\mathrm{\mu }{}_{0}\mathrm{l}{}_{1}\mathrm{l}_{2}}{2\mathrm{\pi r}}$……………(1)

Where ${\mathrm{l}}_{1}\mathrm{and}{\mathrm{l}}_{2}$ is the current in the wires, $\mathrm{r}$ is the separation between the wires, and ${\mathrm{\mu }}_{0}$ is the permeability of free space.

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

$\frac{F}{L}=\frac{4\pi ×{10}^{-7}N/{A}^{2}×{\left(5.00A\right)}^{2}}{2\pi ×\left(3×{10}^{-3}m\right)}\phantom{\rule{0ex}{0ex}}=1.67×{10}^{-3}N/m$

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

## 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.

${F}_{max}=2×\left(1.67×{10}^{-3}N/m\right)\phantom{\rule{0ex}{0ex}}=3.34×{10}^{-3}N/m$

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

## 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. ### Want to see more solutions like these? 