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(B) The resistance of the wire is given by $R = \frac{ho \ell}{A}$.Given $R = 1 \, \Omega$, $ho = 2 	imes 10^{-6} \, \Omega m$, and $A = 10 \, mm

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The resistance of the wire is given by $R = \frac{ho \ell}{A}$.Given $R = 1 \, \Omega$, $ho = 2 	imes 10^{-6} \, \Omega m$, and $A = 10 \, mm^2 = 10 	imes 10^{-6} \, m^2 = 10^{-5} \, m^2$.Substituting these values: $1 = \frac{2 	imes 10^{-6} 	imes \ell}{10^{-5}} \Rightarrow 1 = 0.2 	imes \ell \Rightarrow \ell = 5 \, m$.For the wire to be suspended in mid-air, the magnetic force must balance the gravitational force: $F_m = F_g$.$Bi\ell = mg$.Given $i = 2 \, A$, $m = 500 \, g = 0.5 \, kg$, $g = 10 \, m/s^2$, and $\ell = 5 \, m$.$B 	imes 2 	imes 5 = 0.5 	imes 10$.$10B = 5$.$B = 0.5 \, T = 5 	imes 10^{-1} \, T$.Thus, the magnitude is $5$.

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