Math Problem Statement

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.

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.

Solution

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

B=μ0I2πdB = \frac{\mu_0 I}{2\pi d}

Where:

  • BB is the magnetic field.
  • μ0\mu_0 is the permeability of free space, 4π×10−7 T⋅m/A4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}.
  • II is the current (2.2 A in this case).
  • dd 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 cm=0.075 m7.5 \, \text{cm} = 0.075 \, \text{m} from point A.
  • The current in Wire 1 is I=2.2 AI = 2.2 \, \text{A}.
  • Using the formula for the magnetic field:
= \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. --- 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.

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Math Problem Analysis

Mathematical Concepts

Electromagnetism
Magnetic Fields
Superposition of Magnetic Fields

Formulas

B = (μ₀ * I) / (2π * d)

Theorems

Biot-Savart Law
Superposition of Magnetic Fields

Suitable Grade Level

Grades 11-12 (Physics)

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