Two long parallel wires separated by 0.1 m c | vedclass

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(C) For the third wire to experience no net magnetic force, the magnetic fields produced by the first and second wires at the position of the third wi

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$0.1 \ m$ from the $1^{st}$ wire, away from the $2^{nd}$ wire

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(C) For the third wire to experience no net magnetic force, the magnetic fields produced by the first and second wires at the position of the third wire must be equal in magnitude and opposite in direction.The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Since the currents in the first and second wires are in opposite directions, the magnetic fields produced by them will be in the same direction in the region between them and in opposite directions in the regions outside them.Let the third wire be placed at a distance $x$ from the first wire. If it is placed outside the region between the two wires (on the side of the smaller current, i.e.,the $1 \ A$ wire), the fields will be in opposite directions.Equating the magnitudes of the magnetic fields:$B_1 = B_2$$\frac{\mu_0 I_1}{2 \pi x} = \frac{\mu_0 I_2}{2 \pi (0.1 + x)}$Substituting the given values $I_1 = 1 \ A$ and $I_2 = 2 \ A$:$\frac{1}{x} = \frac{2}{0.1 + x}$$0.1 + x = 2x$$x = 0.1 \ m$Thus, the third wire is placed at a distance of $0.1 \ m$ from the first wire, away from the second wire.

Two long parallel wires separated by $0.1 \ m$ carry currents of $1 \ A$ and $2 \ A$, respectively, in opposite directions. $A$ third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

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