Two very long, straight, parallel wires carry currents that are directed perpendicular to the page as shown in Figure P30.9. Wire 1 carries a current *I*1 into the page (in the -*z* direction) and passes through the *x* axis at *x* = +*a*. Wire 2 passes through the *x* axis at *x* = -2*a* and carries an unknown current *I*2. The total magnetic field at the origin due to the current-carrying wires has the magnitude 2𝜇0*I*1 / ( 2𝜋a ). The current *I*2 can have either of two possible values.

Figure P30.9

(a) Find the value of *I*2 with the smaller magnitude, stating it in terms of *I*1 and giving its direction. (Use the following as necessary: *I*1.)

*I*2 =

out of the page

(b) Find the other possible value of *I*2. (Use the following as necessary: *I*1.)
*I*2 =

---Direction--- to the left to the right upward downward into the page out of the page

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**.

---

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.

Two very long, straight, parallel wires carry currents that are directed perpendicular to the page as shown in Figure P30.9. Wire 1 carries a current I1 into the page (in the -z direction) and passes through the x axis at x = +a. Wire 2 passes through the x axis at x = -2a and carries an unknown current I2. The total magnetic field at the origin due to the current-carrying wires has the magnitude 2μ0I1 / (2πa). The current I2 can have either of two possible values. (a) Find the value of I2 with the smaller magnitude, stating it in terms of I1 and giving its direction. (b) Find the other possible value of I2.

Learn how to find the unknown current I2 in a parallel wire setup, where magnetic fields and current directions influence the total magnetic field at a point. Includes step-by-step calculations, using Biot-Savart Law and superposition for magnetic fields.

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