 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q15PE

Expert-verified Found in: Page 701 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Using the results of the above example on Example 20.3, find the drift velocity in a copper wire of twice the diameter and carrying 20.0 A.

The drift velocity in the copper wire is $1.08 \times {10^{ - 4}}\;{\rm{m/s}}$.

See the step by step solution

## Step 1: Identification of the given data

The given data can be listed below as:

• The gauge of the copper wire is a = 12.
• The diameter of the copper wire is $$d = 2 \times 2.053\;{\rm{mm}} \times \frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{mm}}}}{\rm{ = 4}}{\rm{.106 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}$$.
• The current carried by the copper wire is l = 20.0 A.

## Step 2: Significance of the drift velocity

The drift velocity is illustrated as the average velocity attained by a particle because of the involvement of an electric field. The drift velocity significantly contributes to the current’s change in an electric field.

## Step 3: Determination of the drift velocity

From the example 20.3, the number of copper atoms can be gathered that is$$n = 8.342 \times {10^{28}}\;{{\rm{m}}^{{\rm{ - 3}}}}$$ .

The equation of the magnitude of the current is expressed as:

${v_d} = \frac{I}{{nq\left( {\pi {{\left( {\frac{d}{2}} \right)}^2}} \right)}}$

Here, $$I$$is the magnitude of current, $${v_d}$$ is the drift velocity, $n$ is the number of the copper atoms, $q$ is the charge of one electron and $d$ is the diameter of the copper wire.

Substitute the values in the above equation.

\begin{aligned}{v_d} = \frac{{20.0\;{\rm{A}}}}{{\left( {8.342 \times {{10}^{28}}\;{{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {\left( {3.14} \right){{\left( {\frac{{{\rm{4}}{\rm{.106 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{2}} \right)}^2}} \right)}}\\ = \frac{{20.0\;{\rm{A}}}}{{\left( {1.33 \times {{10}^{10}}\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {\left( {3.14} \right){{\left( {{\rm{2}}{\rm{.106 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}} \right)}^2}} \right)}}\\ = \frac{{20.0\;{\rm{A}}}}{{\left( {1.33 \times {{10}^{10}}\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {\left( {3.14} \right)\left( {{\rm{4}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{{\rm{m}}^2}} \right)} \right)}}\\ = \frac{{20.0\;{\rm{A}}}}{{\left( {1.33 \times {{10}^{10}}\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 3}}}}} \right)\left( {{\rm{1}}{\rm{.39 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{{\rm{m}}^2}} \right)}}\end{aligned}

Hence, further as:

\begin{aligned}{v_d} = \frac{{20.0\;{\rm{A}}}}{{\left( {184870\;{\rm{C}} \cdot {{\rm{m}}^{{\rm{ - 1}}}}} \right)}}\\ = 1.08 \times {10^{ - 4}}\;{\rm{A}} \cdot {\rm{m/C}}\\ = 1.08 \times {10^{ - 4}}\;{\rm{A}} \cdot {\rm{m/C}} \times \frac{{{\rm{1}}\;{\rm{C/s}}}}{{{\rm{1}}\;{\rm{A}}}}\\ = 1.08 \times {10^{ - 4}}\;{\rm{m/s}}\end{aligned}

Thus, the drift velocity in the copper wire is $1.08 \times {10^{ - 4}}\;{\rm{m/s}}$. ### Want to see more solutions like these? 