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

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The resistance of the wire is:$$R = \frac{V}{I} = \frac{1.5 \mathrm{~V}}{8.0 \times 10^{-3} \mathrm{~A}} = 187.5 \Omega$$The resistance of a wire is given by:$$R = \frac{\rho L}{A}$$where:* $\rho$ is the resistivity of the metal* $L$ is the length of the wire* $A$ is the cross-sectional area of the wireSolving for $A$, we get:$$A = \frac{\rho L}{R} = \frac{(2.24 \times 10^{-8} \Omega \cdot \mathrm{m})(1.0 \mathrm{~m})}{187.5 \Omega} = 1.2 \times 10^{-8} \mathrm{~m^2}$$The diameter of the wire is:$$d = 2\sqrt{\frac{A}{\pi}} = 2\sqrt{\frac{1.2 \times 10^{-8} \mathrm{~m^2}}{\pi}} = 12 \mu \mathrm{m}$$

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