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(C) The force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{	ext{eff}} 	imes \vec{B})$,where $\vec{L}_{

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$-0.05$

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(C) The force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{	ext{eff}} 	imes \vec{B})$,where $\vec{L}_{	ext{eff}}$ is the vector displacement from the starting point to the ending point of the wire.From the given figure,the wire starts at $x = -\ell$ and ends at $x = +\ell$ at the height $y = 6 \ m$.Using the equation $x^2 = 6y$,at $y = 6 \ m$,we have $x^2 = 6(6) = 36$,so $x = \pm 6 \ m$.Thus,the effective length vector is $\vec{L}_{	ext{eff}} = (6 - (-6)) \hat{i} = 12 \hat{i} \ m$.The current is $i = 2 \ A$ and the magnetic field is $\vec{B} = 2 	imes 10^{-3} \hat{k} \ T$.Substituting these into the force formula:$\vec{F} = 2 	imes (12 \hat{i} 	imes 2 	imes 10^{-3} \hat{k})$$\vec{F} = 48 	imes 10^{-3} (\hat{i} 	imes \hat{k})$Since $\hat{i} 	imes \hat{k} = -\hat{j}$,we get:$\vec{F} = 48 	imes 10^{-3} (-\hat{j}) = -0.048 \hat{j} \ N$.Rounding this to the nearest value,we get $\vec{F} \simeq -0.05 \hat{j} \ N$.

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