Answer:
2 times the diameter of the original wire.
Explanation:
Let's call [tex]\rho[/tex] the resistivity of the wire. The original resistance of the wire is given by
[tex]R=\rho \frac{L}{A}[/tex]
where L is the length of the wire and A is the cross-sectional area. Writing the area as
[tex]A=\pi (\frac{d}{2})^2[/tex]
where d is the diameter and d/2 corresponds to the radius, we can rewrite the resistance of the wire as
[tex]R=\rho \frac{L}{\pi (d/2)^2}=4\frac{\rho L}{\pi d^2}[/tex]
Which can be rewritten solving for d, the diameter:
[tex]d=\sqrt{\frac{4\rho L}{\pi R}}[/tex]
Now, we know that the wire must be replaced by a wire with same material (so, same resistivity [tex]\rho[/tex]), but with 4 times long, which means the new length is [tex]L'=4L[/tex]. Substituting this into the formula, and keeping in mind that the resistance R must remain the same, we find the new diameter:
[tex]d'=\sqrt{\frac{4 \rho L' }{\pi R}}=\sqrt{\frac{4 \rho (4L) }{\pi R}}=2\sqrt{\frac{4 \rho L' }{\pi R}}=2d[/tex]
so, the diameter must be
2 times the diameter of the original wire.
The diameter of the new wire is 2 times the diameter of the original wire.
The term resistivity has to do with the resistance offered to the flow of current in a circuit.
Given that;
R = ρl/A
We know that the area of the wire = πr^2 or π(d/2)^2
Hence;
R° = 4lρ/ π(d/2)^2
Note that resistance and the resistivity are constant however, the wire is increased to four times its length hence, the diameter of the new wire is 2 times the diameter of the original wire.
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