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(B) The force per unit length on a wire carrying current $I$ due to another parallel wire carrying current $I'$ at a distance $r$ is given by $F = \fr

Vedclass · Accepted Answer

$x = \pm \frac{I_1 d}{(I_1 - I_2)}$

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(B) The force per unit length on a wire carrying current $I$ due to another parallel wire carrying current $I'$ at a distance $r$ is given by $F = \frac{\mu_0 I I'}{2 \pi r}$.For the net force on wire $C$ to be zero,the forces exerted by wires $A$ and $B$ must be equal in magnitude and opposite in direction.Let wire $C$ be at a distance $x$ from $A$ and $(d-x)$ from $B$. The force per unit length due to $A$ is $F_A = \frac{\mu_0 I_1 I}{2 \pi x}$ (repulsive if currents are in the same direction,attractive if opposite).The force per unit length due to $B$ is $F_B = \frac{\mu_0 I_2 I}{2 \pi (d-x)}$.Since the currents $I_1$ and $I_2$ are in opposite directions (as shown in the figure),the forces will be in the same direction if $C$ is between $A$ and $B$. Thus,the equilibrium point must lie outside the region between $A$ and $B$.Equating the magnitudes of the magnetic fields produced by $A$ and $B$ at the position of $C$:$\frac{\mu_0 I_1}{2 \pi x} = \frac{\mu_0 I_2}{2 \pi |x - d|}$$\frac{I_1}{x} = \frac{I_2}{|x - d|}$Case $1$: $x > d$,then $x - d = x - d$,so $I_1(x - d) = I_2 x \Rightarrow x(I_1 - I_2) = I_1 d \Rightarrow x = \frac{I_1 d}{I_1 - I_2}$.Case $2$: $x Re-evaluating the geometry for equilibrium: The point where the net magnetic field is zero is $x = \frac{I_1 d}{I_1 - I_2}$.

Two wires $A$ and $B$ are carrying currents $I_1$ and $I_2$ as shown in the figure. The separation between them is $d$. $A$ third wire $C$ carrying a current $I$ is to be kept parallel to them at a distance $x$ from $A$ such that the net force acting on it is zero. The possible values of $x$ are

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