A particular wire has a diameter of 1.7 mm and length of 1.3 m.If the resistivity of the metal from which the wire is made is 2.0 * 10^-8 $\varOmega$ m,and 45 $\mu$V is the observed potential drop measured across the wire,what current is flowing in the wire?

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Quizplus · Accepted Answer

The resistance of the wire is given by:$$R = \rho \frac{l}{A}$$where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area. The cross-sectional area is given by:$$A = \pi r^2$$where $r$ is the radius. Substituting these expressions into the first equation, we get:$$R = \rho \frac{l}{\pi r^2}$$Plugging in the given values, we get:$$R = (2.0 \times 10^{-8} \Omega \text{m}) \frac{1.3 \text{m}}{\pi (0.85 \times 10^{-3} \text{m})^2} = 0.32 \Omega$$The current is then given by:$$I = \frac{V}{R} = \frac{45 \mu \text{V}}{0.32 \Omega} = 3.9 \text{mA}$$

A Particular Wire Has a Diameter of 1 Ω\varOmegaΩ M,and 45 μ\muμ

A Particular Wire Has a Diameter of 1 $\varOmega$ M,and 45 $\mu$