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Q54PE

Expert-verifiedFound in: Page 813

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 $1.67\times {10}^{3}\mathrm{N}/m$.

(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.

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

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

**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.**

**(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 ${\mathrm{l}}_{1}={\mathrm{l}}_{2}=\mathrm{l}$.

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

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

**(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.

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

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

**(c)**

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

**(d)**

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

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