A straight horizontal long wire carries 30 A | vedclass

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(A) To balance the wire in air,the upward magnetic force must equal the downward gravitational force.The magnetic force on a current-carrying wire is

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$15 	imes 10^{-3} \ T$

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(A) To balance the wire in air,the upward magnetic force must equal the downward gravitational force.The magnetic force on a current-carrying wire is given by $F_m = BI\ell$.The gravitational force is $F_g = mg$.Equating the two: $BI\ell = mg$.Rearranging for the magnetic field $B$: $B = \frac{mg}{I\ell} = \left(\frac{m}{\ell}ight) \frac{g}{I}$.Given linear mass density $\lambda = \frac{m}{\ell} = 45 \ g/m = 45 	imes 10^{-3} \ kg/m$,current $I = 30 \ A$,and taking $g = 10 \ m/s^2$:$B = (45 	imes 10^{-3}) 	imes \frac{10}{30} = 45 	imes 10^{-3} 	imes \frac{1}{3} = 15 	imes 10^{-3} \ T$.

$A$ straight horizontal long wire carries $30 \ A$ current. The linear mass density of the wire is $45 \ g/m$. What is the magnitude of the magnetic field needed to balance the wire in air?

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