Questions: Two long, straight wires are perpendicular to the plane of the paper and at a distance 0.3 m from each other, as shown in the figure. The wires carry currents of I₁=1.5 A and I₂=4.8 A in the direction indicated (out of the page). Find the magnitude and direction of the magnetic field (in μT) at a point A midway between the wires. You need to indicate the direction with a positive or a negative value for the magnetic field. Keep in mind that a vector is positive if directed to the right and negative if directed to the left on the x axis and it is positive if directed up and negative if directed down on the y axis. Your answer should be a number with two decimal places, do not include the unit.

Solution Steps

• The magnetic field \( B \) due to a long straight current-carrying wire at a distance \( r \) is given by \( B = \frac{\mu_0 I}{2 \pi r} \).
• Here, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m}/\text{A} \), \( I_1 = 1.5 \, \text{A} \), \( I_2 = 4.8 \, \text{A} \), and \( r = \frac{d}{2} = 0.15 \, \text{m} \).

• For wire 1: \[ B_1 = \frac{\mu_0 I_1}{2 \pi r} = \frac{4\pi \times 10^{-7} \times 1.5}{2 \pi \times 0.15} = \frac{6 \times 10^{-7}}{0.3} = 2 \times 10^{-6} \, \text{T} = 2 \, \mu\text{T} \]
• For wire 2: \[ B_2 = \frac{\mu_0 I_2}{2 \pi r} = \frac{4\pi \times 10^{-7} \times 4.8}{2 \pi \times 0.15} = \frac{19.2 \times 10^{-7}}{0.3} = 6.4 \times 10^{-6} \, \text{T} = 6.4 \, \mu\text{T} \]

• Using the right-hand rule:
  • For \( I_1 \) (out of the page), the magnetic field at point A is directed downwards (negative y-axis).
  • For \( I_2 \) (into the page), the magnetic field at point A is directed upwards (positive y-axis).

• Since \( B_1 \) is directed downwards and \( B_2 \) is directed upwards: \[ B_{\text{net}} = B_2 - B_1 = 6.4 \, \mu\text{T} - 2 \, \mu\text{T} = 4.4 \, \mu\text{T} \]
• The net magnetic field is directed upwards (positive y-axis).

Final Answer