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(C) The magnetic field $B$ produced by the infinitely long wire $AB$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$. The direction of th

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The rod $CD$ will move upward and turn clockwise at the same time.

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(C) The magnetic field $B$ produced by the infinitely long wire $AB$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$. The direction of this magnetic field is perpendicular to the plane of the paper and directed inwards.For the current-carrying wire $CD$,the magnetic force on a small element $dx$ at distance $r$ from $AB$ is $dF = i_2 (dx) B = i_2 (dx) \frac{\mu_0 i_1}{2\pi r}$.Using Fleming's Left-Hand Rule,the direction of the force on the wire $CD$ is upward (parallel to $AB$).Since the magnetic field $B$ is non-uniform (it decreases as $r$ increases),the force per unit length is greater near $C$ than near $D$. This non-uniform force distribution results in a net upward force and a torque about the center of mass of the rod $CD$. The torque causes the rod to rotate in a clockwise direction as it moves upward.

An infinitely long straight conductor $AB$ is fixed and a current $i_1$ is passed through it. Another movable straight wire $CD$ of finite length carrying current $i_2$ is held perpendicular to it and released. Neglect the weight of the wire.

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