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Question: A copper wire has diameter 0.5mm and resistivity of \[1.6 \times {10^{ - 8}}\,\Omega {\text{m}}\]. W...

A copper wire has diameter 0.5mm and resistivity of 1.6×10−8 Ωm1.6 \times {10^{ - 8}}\,\Omega {\text{m}}. What will be the length of this wire to make its resistance 10 Ω10\,\Omega ?

Explanation

Solution

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=ρlAR = \dfrac{{\rho l}}{A} ,
where, ρ\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 ρ=1.6×10−8 Ωm\rho = 1.6 \times {10^{ - 8}}\,\Omega {\text{m}}, its diameter is
d=0.5 mmd = 0.5\,{\text{mm}} and the resistance is R=10 ΩR = 10\,\Omega .
Let us convert the diameter into the radius of cross section of the wire as follows,
r=d2=0.5 mm2r = \dfrac{d}{2} = \dfrac{{0.5\,{\text{mm}}}}{2}
⇒r=0.25 mm\Rightarrow r = 0.25\,{\text{mm}}
⇒r=0.25×10−3 m\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=πr2A = \pi {r^2}
Substituting r=0.25×10−3 mr = 0.25 \times {10^{ - 3}}\,{\text{m}} in the above equation, we get,
A=(3.14)(0.25×10−3)2A = \left( {3.14} \right){\left( {0.25 \times {{10}^{ - 3}}} \right)^2}
⇒A=1.9625×10−7 m2\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=ρlAR = \dfrac{{\rho l}}{A}
Here, ρ\rho is the resistivity of the material.
Rearranging the above equation for l, we get,
l=RAρl = \dfrac{{RA}}{\rho }
Substituting R=10 ΩR = 10\,\Omega , A=1.9625×10−7 m2A = 1.9625 \times {10^{ - 7}}\,{{\text{m}}^2} and ρ=1.6×10−8 Ωm\rho = 1.6 \times {10^{ - 8}}\,\Omega {\text{m}} in the above equation, we get,
l=(10)(1.9625×10−7)1.6×10−8l = \dfrac{{\left( {10} \right)\left( {1.9625 \times {{10}^{ - 7}}} \right)}}{{1.6 \times {{10}^{ - 8}}}}
⇒l=1.9625×10−61.6×10−8\Rightarrow l = \dfrac{{1.9625 \times {{10}^{ - 6}}}}{{1.6 \times {{10}^{ - 8}}}}
∴l=122 m\therefore l = 122\,{\text{m}}

Thus, the length of wire to have the resistance 10 Ω10\,\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=πd24A = \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.