Two long straight wires are perpendicular to the page and located at distances d1 = 2 cm and d2 = 3.5 cm from the origin. Wire 1 carries current 5 A out of the page. If the net magnetic field due to the two currents is at an angle of 45 degrees to the x-axis at the origin, what is the magnitude of the current in wire 2 [in A]?
To find the magnitude of the current in wire 2, we can use the principle of superposition. The net magnetic field at the origin due to the two currents can be calculated by summing the magnetic fields produced by each wire individually.
The magnetic field produced by a long straight wire at a distance r from the wire is given by the equation:
B = (μ₀ * I) / (2π * r)
Where:
- B is the magnetic field
- μ₀ is the permeability of free space, approximately 4π x 10^-7 (T m/A)
- I is the current in the wire
- r is the distance from the wire to the point where we want to calculate the magnetic field
Let's calculate the magnetic field at the origin due to wire 1:
B1 = (μ₀ * I1) / (2π * d1)
Substituting the values given:
B1 = (4π * 10^-7 T m/A * 5 A) / (2π * 0.02 m)
B1 = 0.5 T
Now, since the magnetic field at the origin is at an angle of 45 degrees to the x-axis, we can decompose it into its x and y components using trigonometry:
Bx = B1 * cos(45°)
By = B1 * sin(45°)
Bx = 0.5 T * cos(45°)
Bx = 0.5 T * (√2/2)
Bx = 0.25 T
By = 0.5 T * sin(45°)
By = 0.5 T * (√2/2)
By = 0.25 T
Now, let's calculate the magnetic field at the origin due to wire 2:
B2 = (μ₀ * I2) / (2π * d2)
Since the net magnetic field at the origin is the vector sum of the magnetic fields due to wire 1 and wire 2, we have:
Bx_total = Bx1 + Bx2
By_total = By1 + By2
Substituting the values we calculated for Bx1 and By1, and the angle of 45 degrees, we have:
Bx_total = 0.25 T + Bx2
By_total = 0.25 T + By2
Since the net magnetic field at the origin is at an angle of 45 degrees to the x-axis, the x and y components of the magnetic field at the origin must be equal:
Bx_total = By_total
Therefore:
0.25 T + Bx2 = 0.25 T + By2
The magnetic field due to wire 2 in both the x and y directions must be equal. Therefore, the magnitude of the current in wire 2 will be the same as the magnitude of the current in wire 1:
I2 = I1 = 5 A
Hence, the magnitude of the current in wire 2 is 5 A.