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(A) For the wire to be suspended in mid-air,the magnetic force acting on it must balance its weight.Let $l$ be the length of the wire,$m$ be its mass,

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) For the wire to be suspended in mid-air,the magnetic force acting on it must balance its weight.Let $l$ be the length of the wire,$m$ be its mass,$I$ be the current,$B$ be the magnetic field,and $g$ be the acceleration due to gravity.The weight of the wire is $W = mg$.Given the linear density $\lambda = \frac{m}{l} = 0.12 \ kg \ m^{-1}$,we can write $m = \lambda l$.So,$W = \lambda l g$.The magnetic force on a current-carrying wire is $F_m = IlB \sin(	heta)$.Since the magnetic field is normal to the wire,$	heta = 90^{\circ}$,so $F_m = IlB$.For equilibrium,$F_m = W$,which implies $IlB = \lambda l g$.Canceling $l$ from both sides,we get $IB = \lambda g$.Solving for $I$: $I = \frac{\lambda g}{B}$.Substituting the given values: $I = \frac{0.12 	imes 10}{0.5} = \frac{1.2}{0.5} = 2.4 \ A$.Therefore,the current through the wire is $2.4 \ A$.

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