An infinitely long,straight conductor AB is f | vedclass

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(C) The magnetic field $B$ produced by the long wire $AB$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2 \pi r}$.According to the right-hand ru

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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 long wire $AB$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2 \pi r}$.According to the right-hand rule,the direction of this magnetic field is into the plane of the paper.The force $dF$ on an element $dx$ of the wire $CD$ at distance $x$ from $AB$ is $dF = i_2 (dx) B = i_2 (dx) \frac{\mu_0 i_1}{2 \pi x}$.Using Fleming's left-hand rule,the direction of this force is upward (away from $AB$ if $i_1$ and $i_2$ are directed as shown).Since the magnetic field $B$ decreases as the distance $x$ increases,the force per unit length is greater near $C$ and smaller near $D$.This non-uniform force distribution results in a net upward force and a torque that causes the rod to rotate clockwise.

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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