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

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(B) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$,where $\vec{L}$ is the vector displacement from the starting point to the end point of the wire.For any current-carrying path between two points $A$ and $C$,the net magnetic force depends only on the straight-line displacement vector $\vec{AC}$ connecting $A$ and $C$.In the given circuit,the current $I$ splits into two paths between $A$ and $C$: path $ABC$ and path $ADC$.For path $ABC$,the displacement vector is $\vec{AC}$ directed from $A$ to $C$ with magnitude $l$. The force is $\vec{F}_{ABC} = I_1 (\vec{l} 	imes \vec{B})$.For path $ADC$,the displacement vector is also $\vec{AC}$ directed from $A$ to $C$ with magnitude $l$. The force is $\vec{F}_{ADC} = I_2 (\vec{l} 	imes \vec{B})$.Since the total current $I = I_1 + I_2$,the total force on the loop is $\vec{F} = \vec{F}_{ABC} + \vec{F}_{ADC} = (I_1 + I_2) (\vec{l} 	imes \vec{B}) = I (\vec{l} 	imes \vec{B})$.Since the magnetic field is perpendicular to the plane of the loop,the magnitude of the force is $F = I l B$.

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