A wire of length L = 6.0 m has a uniform diameter d = 3.0 mm.When an electric field of E = 0.25 V/m is applied to the wire,the current is 1.5 A.The resistivity of the wire is A)1.18 $\times$ 10^-6 $\Omega$.m B)3.53 $\times$ 10^-⁶ $\Omega$.m C)2.81 $\times$ 10^-7 $\Omega$.m D)7.03 $\times$ 10^-⁸ $\Omega$.m E)2.21 $\times$ 10^-7 $\Omega$.m

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The resistivity of the wire is given by:$$ho = \frac{RA}{L}$$where:* $ho$ is the resistivity in ohm-meters* $R$ is the resistance in ohms* $A$ is the cross-sectional area in square meters* $L$ is the length in metersThe resistance of the wire is given by:$$R = \frac{V}{I}$$where:* $V$ is the voltage in volts* $I$ is the current in amperesThe cross-sectional area of the wire is given by:$$A = \frac{\pi d^2}{4}$$where:* $d$ is the diameter in metersSubstituting these equations into the first equation, we get:$$ho = \frac{VLI}{\pi d^2 I}$$$$ho = \frac{V L}{\pi d^2}$$Plugging in the given values, we get:$$ho = \frac{(0.25 	ext{ V})(6.0 	ext{ m})}{\pi (0.003 	ext{ m})^2}$$$$ho = 1.18 	imes 10^{-6} \Omega \cdot 	ext{m}$$Therefore, the correct answer is A.

A Wire of Length L = 6 ×\times× 10-6 Ω\OmegaΩ M B)353 ×\times×

A Wire of Length L = 6 $\times$ 10^-6 $\Omega$ M
B)353 $\times$