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Two parallel wires, each carrying a current of 2.2 A in the same direction, are shown in the figure. Find the magnitude of the net magnetic field at points A.

To solve for the magnitude of the net magnetic field at point \( A \), we need to use the formula for the magnetic field created by a current-carrying wire at a distance from the wire:

\[
B = \frac{\mu_0 I}{2\pi d}
\]

Where:
- \( B \) is the magnetic field.
- \( \mu_0 \) is the permeability of free space, \( 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \).
- \( I \) is the current (2.2 A in this case).
- \( d \) is the distance from the wire to the point where we are calculating the magnetic field.

### Step 1: Magnetic field due to Wire 1 at point A
- Wire 1 is located at a distance \( 7.5 \, \text{cm} = 0.075 \, \text{m} \) from point A.
- The current in Wire 1 is \( I = 2.2 \, \text{A} \).
- Using the formula for the magnetic field:

\[
B_1 = \frac{\mu_0 \times 2.2}{2 \pi \times 0.075}
= \frac{(4 \pi \times 10^{-7}) \times 2.2}{2 \pi \times 0.075}
= \frac{8.8 \times 10^{-7}}{0.15}
= 5.87 \times 10^{-6} \, \text{T}
\]

The direction of the magnetic field generated by Wire 1 at point A can be determined using the right-hand rule. Since the current in Wire 1 is flowing to the right, the magnetic field at point A will be directed into the page.

### Step 2: Magnetic field due to Wire 2 at point A
- The distance from Wire 2 to point A is \( 7.5 \, \text{cm} + 7.5 \, \text{cm} = 15 \, \text{cm} = 0.15 \, \text{m} \).
- The current in Wire 2 is the same, \( I = 2.2 \, \text{A} \).

The magnetic field due to Wire 2 at point A is:

\[
B_2 = \frac{\mu_0 \times 2.2}{2 \pi \times 0.15}
= \frac{(4 \pi \times 10^{-7}) \times 2.2}{2 \pi \times 0.15}
= \frac{8.8 \times 10^{-7}}{0.30}
= 2.93 \times 10^{-6} \, \text{T}
\]

By the right-hand rule, the magnetic field due to Wire 2 at point A will also be directed into the page, as the current flows in the same direction.

### Step 3: Net Magnetic Field at Point A
Since both magnetic fields are in the same direction (into the page), we can simply add them together:

\[
B_{\text{net}} = B_1 + B_2 = 5.87 \times 10^{-6} \, \text{T} + 2.93 \times 10^{-6} \, \text{T}
= 8.8 \times 10^{-6} \, \text{T}
\]

Thus, the magnitude of the net magnetic field at point A is:

\[
B_{\text{net}} = 8.8 \times 10^{-6} \, \text{T} = 8.8 \, \mu\text{T}
\]

Let me know if you want further details or clarifications.

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Here are 5 related questions to consider:
1. How would the net magnetic field change if the current in one of the wires were reversed?
2. What would be the direction of the magnetic field at point C?
3. How would the magnetic field at point B compare to the magnetic fields at points A and C?
4. If the distance between the two wires were doubled, how would that affect the net magnetic field at point A?
5. What if the current in both wires were doubled?

Tip: Use the right-hand rule to determine the direction of the magnetic field when dealing with current-carrying wires.

Two parallel wires, each carrying a current of 2.2 A in the same direction, are shown in the figure. Find the magnitude of the net magnetic field at point A.

This detailed solution explains how to calculate the net magnetic field at point A for two parallel wires carrying currents of 2.2 A each. The solution involves using the Biot-Savart Law and the principle of superposition. Suitable for physics students in Grades 11-12.

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