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san4es73 [151]
3 years ago
13

What length of a certain metal wire of diameter 0.15 mm is needed for the wire to have a resistance of 15 ω? the resistivity of

this metal is 1.68 × 10-8 ω ∙ m?
Physics
1 answer:
Murljashka [212]3 years ago
5 0

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} . Rearranging the above equation,

A=\frac{\rho l}{R}\\  \frac{\pi d^2}{4}=\frac{\rho l}{R}\\  l=\frac{\pi d^2 R}{4 \rho}

Here d=0.15, R=15, \rho=1.68(10^{-8}).

Substituting numerical values,

l=\frac{\pi 0.15^2(10^{-6}) (15)}{4 (1.68)(10^{-8})}  \\ l=15.78

The length of the wire is 15.78 \;\;m

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