A conductor ABCDE,shaped as shown,carries a c | vedclass

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(D) The magnetic force on a current-carrying conductor of any shape is given by $\vec{F} = i(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is t

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(D) The magnetic force on a current-carrying conductor of any shape is given by $\vec{F} = i(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the effective length vector joining the start point $A$ to the end point $E$.From the figure,the distance between $A$ and $E$ is $\lambda$ along the $x$-axis. Thus,$\vec{L}_{eff} = \lambda \hat{i}$.$1$. If $\vec{B} = B\hat{i}$,then $\vec{F} = i(\lambda \hat{i} 	imes B\hat{i}) = 0$. (Option $A$ is correct)$2$. If $\vec{B} = B\hat{j}$,then $\vec{F} = i(\lambda \hat{i} 	imes B\hat{j}) = \lambda Bi \hat{k}$. (Option $B$ is correct)$3$. If $\vec{B} = B\hat{k}$,then $\vec{F} = i(\lambda \hat{i} 	imes B\hat{k}) = -\lambda Bi \hat{j}$. (Option $C$ is correct)Since all statements are correct,the final answer is $D$.

$A$ conductor $ABCDE$,shaped as shown,carries a current $i$. It is placed in the $xy$ plane with the ends $A$ and $E$ on the $x$-axis. $A$ uniform magnetic field of magnitude $B$ exists in the region. The force acting on it will be

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