In the figure shown,a current I_1 is establis | vedclass

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(B) The magnetic field $B$ produced by the long straight wire $AB$ at any point in the plane of the paper is directed perpendicularly inwards (into th

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towards left

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(B) The magnetic field $B$ produced by the long straight wire $AB$ at any point in the plane of the paper is directed perpendicularly inwards (into the paper) according to the Right-Hand Thumb Rule.For any small current element $Idl$ on the wire $CD$,the force is given by $dF = I(dl 	imes B)$.Since the magnetic field $B$ is uniform in direction (perpendicular to the plane) and the current $I_2$ flows through the wire $CD$,we can consider the net force on the wire $CD$ as the vector sum of forces on its segments.Due to the symmetry of the wire $CD$ with respect to the magnetic field produced by $AB$,the horizontal components of the forces on the two segments of $CD$ cancel each other out.The vertical components of the forces on both segments of $CD$ are directed towards the wire $AB$ (leftwards) because the current in both segments of $CD$ has a component parallel to $AB$ or is oriented such that the cross product $dl 	imes B$ results in a force towards the left.Therefore,the net force on the wire $CD$ is directed towards the left.

In the figure shown,a current $I_1$ is established in the long straight wire $AB$. Another wire $CD$ carrying current $I_2$ is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire $AB$. The force on the wire $CD$ is:

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