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(C) The current in the circuit is $I = \frac{V}{R} = \frac{24}{12} = 2 \, A$.The initial equilibrium condition for the wire of mass $m$ supported by t

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$0.6 \, T$ and directed into the plane of the page

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(C) The current in the circuit is $I = \frac{V}{R} = \frac{24}{12} = 2 \, A$.The initial equilibrium condition for the wire of mass $m$ supported by two springs of constant $k$ is $2kx_1 = mg$,where $x_1 = 0.5 \, cm = 0.5 	imes 10^{-2} \, m$.When the magnetic field $B$ is applied,the additional force $F_m = IlB$ causes an additional stretch $x_2 = 0.3 \, cm = 0.3 	imes 10^{-2} \, m$. The new equilibrium condition is $2k(x_1 + x_2) = mg + IlB$.Subtracting the first equation from the second gives $2kx_2 = IlB$.Substituting $2k = \frac{mg}{x_1}$ into the equation,we get $\frac{mg}{x_1} x_2 = IlB$.Solving for $B$: $B = \frac{mgx_2}{Ilx_1} = \frac{10 	imes 10^{-3} 	imes 10 	imes 0.3 	imes 10^{-2}}{2 	imes 5 	imes 10^{-2} 	imes 0.5 	imes 10^{-2}} = \frac{0.3}{0.5} = 0.6 \, T$.Since the springs stretch further,the magnetic force must be directed downwards. By Fleming's Left-Hand Rule,for a current flowing in the wire,the magnetic field must be directed into the plane of the page to produce a downward force.

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