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Q15PE

Expert-verifiedFound in: Page 701

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 \[1.08 \times {10^{ - 4}}\;{\rm{m/s}}\].

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.

**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.**

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.

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

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 \[1.08 \times {10^{ - 4}}\;{\rm{m/s}}\].

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