A wire 50 cm long and 1 mm^2 in cross-secti | vedclass

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(B) Given: Length $\ell = 50 \ cm = 0.5 \ m$,Area $A = 1 \ mm^2 = 10^{-6} \ m^2$,Voltage $V = 2 \ V$,Current $I = 4 \ A$.First,calculate the resistanc

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$1 	imes 10^{-6} \ \Omega \cdot m$

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(B) Given: Length $\ell = 50 \ cm = 0.5 \ m$,Area $A = 1 \ mm^2 = 10^{-6} \ m^2$,Voltage $V = 2 \ V$,Current $I = 4 \ A$.First,calculate the resistance $R$ using Ohm's Law: $R = \frac{V}{I} = \frac{2}{4} = 0.5 \ \Omega$.The formula for resistivity $ho$ is given by $R = ho \frac{\ell}{A}$,which implies $ho = \frac{R \cdot A}{\ell}$.Substituting the values: $ho = \frac{0.5 	imes 10^{-6}}{0.5} = 1 	imes 10^{-6} \ \Omega \cdot m$.

$A$ wire $50 \ cm$ long and $1 \ mm^2$ in cross-section carries a current of $4 \ A$ when connected to a $2 \ V$ battery. The resistivity of the wire is:

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