When a 1.0-m length of metal wire is connected to a $1.5 - \mathrm { V }$ battery, a current of $8.0 \mathrm {~mA}$ flows through it. What is the diameter of the wire? The resistivity of the metal is $2.24 \times 10 ^ { - 8 } \Omega \cdot \mathrm { m }$.
A) $12 \mu \mathrm { m }$
B) $24 \mu \mathrm { m }$
C) $6.0 \mu \mathrm { m }$
D) $2.2 \mu \mathrm { m }$

Question

Quizplus · Accepted Answer

The resistance of the wire is given by:$$R = \frac{\rho L}{A}$$where $\rho$ is the resistivity of the metal, $L$ is the length of the wire, and $A$ is the cross-sectional area of the wire.The current through the wire is given by:$$I = \frac{V}{R}$$where $V$ is the voltage of the battery.Substituting the given values into these equations, we get:$$R = \frac{(2.24 \times 10^{-8} \Omega \cdot \mathrm{m})(1.0 \mathrm{m})}{\pi r^2}$$$$I = \frac{1.5 \mathrm{V}}{R}$$Solving for $r$, we get:$$r = \sqrt{\frac{\rho L}{\pi I}} = \sqrt{\frac{(2.24 \times 10^{-8} \Omega \cdot \mathrm{m})(1.0 \mathrm{m})}{\pi (8.0 \times 10^{-3} \mathrm{A})}} = 6.0 \mu \mathrm{m}$$Therefore, the diameter of the wire is $2r = 12 \mu \mathrm{m}$.

When a 10-M Length of Metal Wire Is Connected to A to A 1.5