The wire in the drawing carries a current of 13 A. Suppose that a second long, straight wire is placed right next to this wire. The current in the second wire is 28 A. Use Ampere's law to find the magnitude of the magnetic field at a distance of r = 0.76 m from the wires when the currents are (a) in the same direction and (b) in opposite directions.
Problem #60
http://drjj.uitm.edu.my/DRJJ/Lecture/PHY407/Sample%20problems%20Chap%2021%20Cutnell.pdf
To find the magnitude of the magnetic field at a distance of r = 0.76 m from the wires using Ampere's law, follow these steps:
1. Determine the path along which we'll apply Ampere's law. In this case, we can choose a circular path centered on the current-carrying wires, with a radius of r = 0.76 m.
2. Ampere's law states that the line integral of the magnetic field B along a closed path is equal to the product of the magnetic permeability of free space (μ₀) and the total enclosed current (I_enc). Since we have two wires, we need to consider the contribution from both wires separately.
3. Calculate the total enclosed current (I_enc) for each case:
a) When the currents in both wires are in the same direction:
- For the first wire, I₁ = 13 A. Since the chosen path encloses only this wire, the total enclosed current is simply I_enc₁ = 13 A.
- For the second wire, I₂ = 28 A. As this wire is outside the chosen path, it does not contribute to the total enclosed current, so I_enc₂ = 0 A.
b) When the currents in the wires are in opposite directions:
- For the first wire, I₁ = 13 A. Again, this wire is the only one enclosed by the chosen path, so I_enc₁ = 13 A.
- For the second wire, I₂ = -28 A. Since the current is in the opposite direction, it is subtracted from the total enclosed current, so I_enc₂ = -28 A.
4. Using Ampere's law, the formula for the magnetic field B at a distance r from the wires is given by:
B = (μ₀ * I_enc) / (2πr)
5. Plug in the values for μ₀, I_enc, and r, and calculate the magnetic field for each case.
a) When the currents are in the same direction:
- μ₀ = 4π × 10⁻⁷ T m/A (magnetic permeability of free space)
- I_enc = I_enc₁ = 13 A (total enclosed current for the first wire)
- r = 0.76 m (distance from the wires)
Substituting these values into the formula:
B = (4π × 10⁻⁷ T m/A * 13 A) / (2π * 0.76 m)
= (52π × 10⁻⁶ T m) / (1.52π m)
≈ 34.21 × 10⁻⁶ T
≈ 34.21 μT (microteslas)
b) When the currents are in opposite directions:
- μ₀ = 4π × 10⁻⁷ T m/A (magnetic permeability of free space)
- I_enc = I_enc₁ + I_enc₂ = 13 A + (-28 A) (total enclosed current for both wires)
- r = 0.76 m (distance from the wires)
Substituting these values into the formula:
B = (4π × 10⁻⁷ T m/A * (-15 A)) / (2π * 0.76 m)
= (-60π × 10⁻⁶ T m) / (1.52π m)
= -39.47 × 10⁻⁶ T
≈ -39.47 μT (microteslas)
Therefore, the magnitude of the magnetic field at a distance of 0.76 m from the wires is approximately 34.21 μT when the currents are in the same direction and -39.47 μT when the currents are in opposite directions.