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(A) The magnetic 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}_{

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$(A)$ If $\vec{B}$ is along $\hat{z}$,$F \propto (L+R)$

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(A) The magnetic 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 vector displacement from the starting point to the end point of the conductor.Looking at the geometry,the conductor starts at some point and ends at a point displaced by a total horizontal distance of $L + R + R + L = 2(L+R)$ along the $x$-axis.Thus,$\vec{L}_{eff} = 2(L+R)\hat{i}$.Therefore,$\vec{F} = I(2(L+R)\hat{i} 	imes \vec{B}) = 2I(L+R)(\hat{i} 	imes \vec{B})$.$(A)$ If $\vec{B} = B\hat{z}$,then $\vec{F} = 2I(L+R)B(\hat{i} 	imes \hat{z}) = -2I(L+R)B\hat{j}$. The magnitude $F = 2I(L+R)B$,so $F \propto (L+R)$. This is correct.$(B)$ If $\vec{B} = B\hat{x}$,then $\vec{F} = 2I(L+R)B(\hat{i} 	imes \hat{x}) = 0$. This is correct.$(C)$ If $\vec{B} = B\hat{y}$,then $\vec{F} = 2I(L+R)B(\hat{i} 	imes \hat{y}) = 2I(L+R)B\hat{z}$. The magnitude $F = 2I(L+R)B$,so $F \propto (L+R)$. This is correct.$(D)$ If $\vec{B} = B\hat{z}$,$F 
eq 0$. This is incorrect.Thus,the correct statements are $(A), (B),$ and $(C)$.

$A$ conductor (shown in the figure) carrying constant current $I$ is kept in the $x-y$ plane in a uniform magnetic field $\vec{B}$. If $F$ is the magnitude of the total magnetic force acting on the conductor,then the correct statement$(s)$ is(are):

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