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(B) The magnetic field $B$ at a distance $x$ from an infinitely long wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi x}$.As the curren

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move towards the wire

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(B) The magnetic field $B$ at a distance $x$ from an infinitely long wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi x}$.As the current $I$ in the wire is suddenly halved, the magnetic flux $\phi$ linked with the square loop decreases.According to Lenz's law, the induced current in the loop will flow in such a direction as to oppose this decrease in magnetic flux. This means the induced current will create a magnetic field in the same direction as the original magnetic field.For this to happen, the induced current in the loop must flow in the clockwise direction.Now, consider the forces on the sides of the loop:$1$. The side of the loop closer to the wire (let's call it $CD$) carries current in the same direction as the main wire. Thus, it experiences an attractive force $F_{CD}$ towards the wire.$2$. The side of the loop farther from the wire (let's call it $AB$) carries current in the opposite direction to the main wire. Thus, it experiences a repulsive force $F_{AB}$ away from the wire.Since the magnetic field is stronger closer to the wire, the attractive force on the closer side is greater than the repulsive force on the farther side $(F_{CD} > F_{AB})$.Therefore, the net force on the loop is directed towards the wire, and the loop will move towards the wire.

$A$ square conducting loop is placed near an infinitely long current-carrying wire with one edge parallel to the wire as shown in the figure. If the current in the straight wire is suddenly halved, which of the following statements will be true? "The loop will .......".

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