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A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current and is also at 20.0°C. Let Ecu be the electric field in the copper wire and Eag be the electric field in the silver wire. Use the resistivities at room temperature for copper. PCu = 1.72 x 10^-8.m and for silver PAE= 1.47 x 10^-8.m. What is the ratio Ecu/Eng?

Answer :

princessesther2011

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

Lanuel

The ratio of the electric field in the copper wire to the electric field in the silver wire is 0.457.

Given the following data:

  • Radius of copper wire = 0.800 mm
  • Temperature = 20.0°C
  • Radius of silver wire = 0.500 mm
  • Resistivity for copper = [tex]1.72 \times 10^{-8}\;m[/tex]
  • Resistivity for silver = [tex]1.47 \times 10^{-8}\;m[/tex]

Conversion:

Radius of copper wire = 0.800 mm to m = 0.0008 meter.

Radius of silver wire = 0.500 mm to m = 0.0005 meter.

To find the ratio of the electric field in the copper wire to the electric field in the silver wire:

First of all, we would determine the area of copper and silver wire respectively.

For copper wire:

[tex]Area = \pi r^2\\\\A_{Cu} = \pi \times (0.0008)^2\\\\A_{Cu} = \pi \times 6.4 \times 10^{-7}\\\\A_{Cu} = 6.4 \times 10^{-7}\pi\;m^2[/tex]

For silver wire:

[tex]Area = \pi r^2\\\\A_{Ag} = \pi \times (0.0005)^2\\\\A_{Ag} = \pi \times 2.5 \times 10^{-7}\\\\A_{Ag} = 2.5 \times 10^{-7}\pi\;m^2[/tex]

Next, we would determine the electric field in each:

Mathematically, electric field is given by the formula:

[tex]E = \frac{PI}{A}[/tex]

For copper wire:

[tex]E_{Cu} = \frac{P_{Cu} I}{A_{Cu} } \\\\E_{Cu} = \frac{1.72 \times 10^{-8}I}{6.4 \times 10^{-7}\pi}[/tex]

For silver wire:

[tex]E_{Ag} = \frac{P_{Ag} I}{A_{Ag} } \\\\E_{Ag} = \frac{1.47 \times 10^{-8}I}{2.5 \times 10^{-7}\pi}[/tex]

Now, we can find the ratio:

[tex]{\frac{E_{Cu}}{E_{Ag}} = \frac{\frac{1.72 \times 10^{-8}I}{6.4 \times 10^{-7}\pi}}{\frac{1.47 \times 10^{-8}I}{2.5 \times 10^{-7}\pi}} }[/tex]

[tex]\frac{E_{Cu}}{E_{Ag}} = \frac{1.72 \times 10^{-8}I}{6.4 \times 10^{-7}\pi} \times \frac{2.5 \times 10^{-7}\pi}{1.47 \times 10^{-8}I} \\\\\frac{E_{Cu}}{E_{Ag}} = \frac{1.72 \times 10^{-8} \;\times \;2.5 \times 10^{-7}}{6.4 \times 10^{-7}\; \times \;1.47 \times 10^{-8}} \\\\\frac{E_{Cu}}{E_{Ag}} = \frac{4.3 \times 10^{-15}}{9.41 \times 10^{-15}} \\\\\frac{E_{Cu}}{E_{Ag}} = 0.457[/tex]

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