Two wires A & 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:

Step-by-Step Solution

1. Condition for Zero Net Force on Wire C

For a third parallel current-carrying wire $$C$$ to experience a net magnetic force of zero, the magnetic fields produced by wire $$A$$ ($$B_1$$) and wire $$B$$ ($$B_2$$) must be equal in magnitude but opposite in direction at the position of wire $$C$$.

$$F_{\text{net}} = 0 \implies B_1 = B_2$$

The magnetic field at a perpendicular distance $$r$$ from a long straight wire carrying current $$I$$ is given by:

$$B = \frac{\mu_0 I}{2\pi r}$$

2. Analysis of Directonal Fields (Regions)

Let's analyze the directions of the currents:

• Wire $$A$$ carries current $$I_1$$ upwards.
• Wire $$B$$ carries current $$I_2$$ downwards (opposite directions).
• Between the wires ($$0 < x < d$$): According to the right-hand grip rule, both wires produce magnetic fields pointing into (or out of) the page in the space between them. Since the fields point in the same direction, they reinforce each other, and the net field can never be zero in this region.
• Outside the wires ($$x < 0$$ or $$x > d$$): The magnetic fields point in opposite directions. Therefore, the net force can only be zero outside the two wires.
• $$x = \frac{I_1 d}{I_2 - I_1}$$ to the left of wire A (if $$I_2 > I_1$$)
• $$x = \frac{I_1 d}{I_1 - I_2}$$ to the right of wire B (if $$I_1 > I_2$$)

Because the currents are in opposite directions:

The exact valid region depends entirely on which current has a smaller magnitude, as wire $$C$$ must be placed closer to the weaker current to balance the stronger current's field.

3. Setting Up the Balance Equations

Case 1: Wire C is placed to the left of Wire A ($$x$$ is measured from A to the left)

Let $$x$$ be the distance from wire $$A$$ to the left. The distance from wire $$B$$ becomes $$(d + x)$$.

Equating the field magnitudes:

$$\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi (d + x)}$$

Cancel out the common constant factors $$\frac{\mu_0}{2\pi}$$:

$$\frac{I_1}{x} = \frac{I_2}{d + x}$$

Cross-multiply and solve for $$x$$:

$$I_1(d + x) = I_2 x$$

$$I_1 d + I_1 x = I_2 x$$

$$I_1 d = (I_2 - I_1)x$$

$$x = \frac{I_1 d}{I_2 - I_1}$$

(This solution is physically valid only if $$I_2 > I_1$$.)

Case 2: Wire C is placed to the right of Wire B ($$x$$ is measured from A to the right)

If wire $$C$$ is placed to the right of wire $$B$$ at a distance $$x$$ from wire $$A$$, its distance from wire $$B$$ is $$(x - d)$$.

Equating the fields:

$$\frac{\mu_0 I_1}{2\pi x} = \frac{\mu_0 I_2}{2\pi (x - d)} \implies \frac{I_1}{x} = \frac{I_2}{x - d}$$

Cross-multiply and solve for $$x$$:

$$I_1(x - d) = I_2 x$$

$$I_1 x - I_1 d = I_2 x$$

$$(I_1 - I_2)x = I_1 d$$

$$x = \frac{I_1 d}{I_1 - I_2}$$

(This solution is physically valid only if $$I_1 > I_2$$.)

Final Answer

The possible values of $$x$$ depend on the relative magnitudes of the currents: