Let's tackle this step by step, focusing on the magnetic fields created by each wire and the conditions given.

Background Information

The magnetic field due to a long, straight current-carrying wire at a point distance $r$ from the wire is given by:

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

where:

• $\mu_0$ is the permeability of free space,
• $I$ is the current in the wire,
• $r$ is the distance from the wire.

The direction of the magnetic field around a current-carrying wire follows the right-hand rule.

Problem Setup

1. Wire 1 is located at $x = +a$, carrying current $I_1$ into the page.
2. Wire 2 is located at $x = -2a$, carrying current $I_2$ (unknown) in some direction.

The total magnetic field at the origin due to both wires has a magnitude of:

$B_{\text{total}} = \frac{2 \mu_0 I_1}{2 \pi a}$

Since the two wires are at different distances from the origin, the magnetic field contributions will differ. Wire 1 is at distance $a$, while Wire 2 is at distance $2a$ from the origin.

Magnetic Field Contribution by Each Wire

• Magnetic Field from Wire 1 at the Origin: $B_1 = \frac{\mu_0 I_1}{2 \pi a}$
• Magnetic Field from Wire 2 at the Origin: $B_2 = \frac{\mu_0 I_2}{2 \pi (2a)} = \frac{\mu_0 I_2}{4 \pi a}$

Conditions on the Magnetic Fields

For the total magnetic field at the origin, we know:

$B_{\text{total}} = B_1 \pm B_2 = \frac{2 \mu_0 I_1}{2 \pi a}$

Solving for $I_2$

1. Case 1 (Fields Add): $B_1 + B_2 = \frac{2 \mu_0 I_1}{2 \pi a}$ Substituting the expressions for $B_1$ and $B_2$: $\frac{\mu_0 I_1}{2 \pi a} + \frac{\mu_0 I_2}{4 \pi a} = \frac{2 \mu_0 I_1}{2 \pi a}$ Simplifying and solving for $I_2$: $\frac{\mu_0 I_2}{4 \pi a} = \frac{\mu_0 I_1}{2 \pi a}$ $I_2 = 2 I_1$

2. Case 2 (Fields Subtract): $B_1 - B_2 = \frac{2 \mu_0 I_1}{2 \pi a}$ $\frac{\mu_0 I_1}{2 \pi a} - \frac{\mu_0 I_2}{4 \pi a} = \frac{2 \mu_0 I_1}{2 \pi a}$ Solving for $I_2$: $-\frac{\mu_0 I_2}{4 \pi a} = \frac{\mu_0 I_1}{2 \pi a}$ $I_2 = -2 I_1$

Answer

(a) The value of $I_2$ with the smaller magnitude is:

$I_2 = -\frac{I_1}{2}$

The direction of this current $I_2$ will be out of the page.

(b) The other possible value of $I_2$ is:

$I_2 = 2 I_1$

The direction of this current $I_2$ will be into the page.

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Let me know if you would like further clarification on any step or additional questions related to this problem. Here are some questions for deeper exploration:

1. How does the right-hand rule apply to determine the direction of the magnetic field?
2. What happens to the magnetic field at the origin if $I_1$ changes direction?
3. How would the solution change if the wires were not infinitely long?
4. What is the effect of increasing the separation between the wires on the magnetic field at the origin?
5. How would the calculation change if the wires were oriented differently?

Tip: Remember that magnetic fields from parallel wires with currents in opposite directions tend to cancel each other out at points equidistant from the wires.