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

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$F = 2B_0I$,upwards

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(A) The force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the displacement vector from the starting point to the ending point of the wire.The equation of the wire is $y = 2x - x^2$.To find the endpoints,set $y = 0$:$0 = 2x - x^2 = x(2 - x)$This gives $x = 0$ and $x = 2$.So,the wire starts at $(0, 0)$ and ends at $(2, 0)$.The effective length vector is $\vec{L}_{eff} = (2 - 0)\hat{i} = 2\hat{i}$.The magnetic field is $\vec{B} = -B_0 \hat{k}$.Now,calculate the force:$\vec{F} = I(2\hat{i} 	imes -B_0 \hat{k})$$\vec{F} = -2B_0I (\hat{i} 	imes \hat{k})$Since $\hat{i} 	imes \hat{k} = -\hat{j}$,we have:$\vec{F} = -2B_0I (-\hat{j}) = 2B_0I \hat{j}$.Since the result is in the $+\hat{j}$ direction,the force is upwards.

$A$ wire carrying a current $I$ is placed inside a uniform magnetic field $\vec{B} = -B_0 \hat{k}$. The shape of the wire is parabolic and has the equation $y = 2x - x^2$. The force on the wire will be:

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