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(B) The magnetic field $\vec{B}$ is directed from the North pole $(N)$ to the South pole $(S)$,which is from right to left (along the negative $x$-axi

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$Q$

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(B) The magnetic field $\vec{B}$ is directed from the North pole $(N)$ to the South pole $(S)$,which is from right to left (along the negative $x$-axis,i.e.,$-\hat{i}$).The current $i$ flows out of the plane of the paper (along the positive $z$-axis,i.e.,$+\hat{k}$).The magnetic force $\vec{F}$ on a current-carrying conductor is given by $\vec{F} = i(\vec{l} 	imes \vec{B})$.Here,$\vec{l}$ is in the direction of current $(+\hat{k})$ and $\vec{B} = -B\hat{i}$.Substituting these values: $\vec{F} = i(l\hat{k} 	imes -B\hat{i}) = -ilB(\hat{k} 	imes \hat{i}) = -ilB\hat{j}$.Since $+\hat{j}$ is towards $P$,the direction $-\hat{j}$ is towards $Q$.Therefore,the conductor will experience a force in the direction towards $Q$.

$A$ straight current-carrying conductor is placed such that the current flows out of the plane of the paper. The conductor is placed between the poles of two magnets,as shown in the figure. The conductor will experience a force in the direction towards:

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