
A copper wire has diameter 0.5mm and resistivity of \[1.6 \times {10^{ - 8}}\,\Omega {\text{m}}\]. What will be the length of this wire to make its resistance \[10\,\Omega \]?
Answer
547.2k+ views
Hint: Determine the radius of the wire since we have given the diameter. The resistance of the wire is proportional to its length and it is inversely proportional to the area of cross section. Rearranging the equation for the length of wire and substituting the given values, you can calculate the required length of the wire.
Formula used:
Resistance, \[R = \dfrac{{\rho l}}{A}\] ,
where, \[\rho \] is the resistivity of the material, l is the length and A is the area of cross-section.
Complete step by step answer:
We have given that the resistivity of the copper is \[\rho = 1.6 \times {10^{ - 8}}\,\Omega {\text{m}}\], its diameter is
\[d = 0.5\,{\text{mm}}\] and the resistance is \[R = 10\,\Omega \].
Let us convert the diameter into the radius of cross section of the wire as follows,
\[r = \dfrac{d}{2} = \dfrac{{0.5\,{\text{mm}}}}{2}\]
\[ \Rightarrow r = 0.25\,{\text{mm}}\]
\[ \Rightarrow r = 0.25 \times {10^{ - 3}}\,{\text{m}}\]
We know that the area of cross section of the wire of the cross-sectional radius r is,
\[A = \pi {r^2}\]
Substituting \[r = 0.25 \times {10^{ - 3}}\,{\text{m}}\] in the above equation, we get,
\[A = \left( {3.14} \right){\left( {0.25 \times {{10}^{ - 3}}} \right)^2}\]
\[ \Rightarrow A = 1.9625 \times {10^{ - 7}}\,{{\text{m}}^2}\]
We have the formula for the resistance of the wire of length l and area of cross section A,
\[R = \dfrac{{\rho l}}{A}\]
Here, \[\rho \] is the resistivity of the material.
Rearranging the above equation for l, we get,
\[l = \dfrac{{RA}}{\rho }\]
Substituting \[R = 10\,\Omega \], \[A = 1.9625 \times {10^{ - 7}}\,{{\text{m}}^2}\] and \[\rho = 1.6 \times {10^{ - 8}}\,\Omega {\text{m}}\] in the above equation, we get,
\[l = \dfrac{{\left( {10} \right)\left( {1.9625 \times {{10}^{ - 7}}} \right)}}{{1.6 \times {{10}^{ - 8}}}}\]
\[ \Rightarrow l = \dfrac{{1.9625 \times {{10}^{ - 6}}}}{{1.6 \times {{10}^{ - 8}}}}\]
\[ \therefore l = 122\,{\text{m}}\]
Thus, the length of wire to have the resistance \[10\,\Omega \] should be 122 m.
Note: If instead of resistivity, the conductivity is given, then students must know that the resistivity is the reciprocal of conductivity. The resistance of the wire decreases as the area of the cross section increases and it increases as the length of the wire increases. Students can also express the area of cross section as \[A = \dfrac{{\pi {d^2}}}{4}\], where, d is the diameter of the wire, so that you can avoid calculating the radius of the cross section of wire.
Formula used:
Resistance, \[R = \dfrac{{\rho l}}{A}\] ,
where, \[\rho \] is the resistivity of the material, l is the length and A is the area of cross-section.
Complete step by step answer:
We have given that the resistivity of the copper is \[\rho = 1.6 \times {10^{ - 8}}\,\Omega {\text{m}}\], its diameter is
\[d = 0.5\,{\text{mm}}\] and the resistance is \[R = 10\,\Omega \].
Let us convert the diameter into the radius of cross section of the wire as follows,
\[r = \dfrac{d}{2} = \dfrac{{0.5\,{\text{mm}}}}{2}\]
\[ \Rightarrow r = 0.25\,{\text{mm}}\]
\[ \Rightarrow r = 0.25 \times {10^{ - 3}}\,{\text{m}}\]
We know that the area of cross section of the wire of the cross-sectional radius r is,
\[A = \pi {r^2}\]
Substituting \[r = 0.25 \times {10^{ - 3}}\,{\text{m}}\] in the above equation, we get,
\[A = \left( {3.14} \right){\left( {0.25 \times {{10}^{ - 3}}} \right)^2}\]
\[ \Rightarrow A = 1.9625 \times {10^{ - 7}}\,{{\text{m}}^2}\]
We have the formula for the resistance of the wire of length l and area of cross section A,
\[R = \dfrac{{\rho l}}{A}\]
Here, \[\rho \] is the resistivity of the material.
Rearranging the above equation for l, we get,
\[l = \dfrac{{RA}}{\rho }\]
Substituting \[R = 10\,\Omega \], \[A = 1.9625 \times {10^{ - 7}}\,{{\text{m}}^2}\] and \[\rho = 1.6 \times {10^{ - 8}}\,\Omega {\text{m}}\] in the above equation, we get,
\[l = \dfrac{{\left( {10} \right)\left( {1.9625 \times {{10}^{ - 7}}} \right)}}{{1.6 \times {{10}^{ - 8}}}}\]
\[ \Rightarrow l = \dfrac{{1.9625 \times {{10}^{ - 6}}}}{{1.6 \times {{10}^{ - 8}}}}\]
\[ \therefore l = 122\,{\text{m}}\]
Thus, the length of wire to have the resistance \[10\,\Omega \] should be 122 m.
Note: If instead of resistivity, the conductivity is given, then students must know that the resistivity is the reciprocal of conductivity. The resistance of the wire decreases as the area of the cross section increases and it increases as the length of the wire increases. Students can also express the area of cross section as \[A = \dfrac{{\pi {d^2}}}{4}\], where, d is the diameter of the wire, so that you can avoid calculating the radius of the cross section of wire.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

