Three long, thin, co-planar wires are separated by 10cm and arranged as shown. The top wire carries 1A, the middle wire carries 2A, and the bottom wire carries 3A.
2A (6 points possible)
What is the magnitude of the force per unit length on the TOP wire in N/m?
1000 - incorrect
1000
What is the direction of the force on the TOP wire?
left right up up - incorrect down into page out of page
You have used 3 of 3 submissions
2B (6 points possible)
What is the magnitude of the force per unit length on the MIDDLE wire in N/m?
What is the magnitude of the force per unit length on the MIDDLE wire in N/m? - unanswered
What is the direction of the force on the MIDDLE wire?
- unanswered left right up down into page out of page
Check your answer Save your answer You have used 0 of 3 submissions
2C (6 points possible)
What is the magnitude of the force on the BOTTOM wire in N/m?
What is the magnitude of the force on the BOTTOM wire in N/m? - unanswered
What is the direction of the force on the bottom wire?
- unanswered left right up down into page out of page
Check your answer Save your answer You have used 0 of 3 submissions
To find the force per unit length on a wire due to the magnetic field created by other wires, we can use the formula:
F = (μ₀ * I₁ * I₂ * L) / (2π * d)
Where:
F is the force per unit length in N/m,
μ₀ is the permeability of free space (4π × 10⁻⁷ T m/A),
I₁ and I₂ are the currents in the two wires in Amperes,
L is the length of the interacting wires in meters, and
d is the distance between the wires in meters.
Let's calculate the force per unit length on each wire.
For the top wire:
I₁ = 1A, I₂ = 2A, L = length of the wire, and d = 10cm = 0.1m.
F_top = (4π × 10⁻⁷ * 1 * 2 * L) / (2π * 0.1) = (8π × 10⁻⁷ * L) / 0.1 = 80π × 10⁻⁷ * L N/m
Therefore, the magnitude of the force per unit length on the top wire is 80π × 10⁻⁷ * L N/m.
Now, for the direction of the force, we can use the right-hand grip rule. Wrap your right-hand fingers around the top wire in the direction of the current (from current source towards the wire), and your thumb will point in the direction of the force.
From the given options, the correct direction of the force on the top wire is "out of the page."
Next, let's move on to the middle wire.
For the middle wire:
I₁ = 2A, I₂ = 3A, L = length of the wire, and d = 10cm = 0.1m.
F_middle = (4π × 10⁻⁷ * 2 * 3 * L) / (2π * 0.1) = (24π × 10⁻⁷ * L) / 0.1 = 240π × 10⁻⁷ * L N/m
Therefore, the magnitude of the force per unit length on the middle wire is 240π × 10⁻⁷ * L N/m.
Using the right-hand grip rule, the correct direction of the force on the middle wire is "into the page."
Finally, let's find the force per unit length on the bottom wire.
For the bottom wire:
I₁ = 1A, I₂ = 3A, L = length of the wire, and d = 10cm = 0.1m.
F_bottom = (4π × 10⁻⁷ * 1 * 3 * L) / (2π * 0.1) = (12π × 10⁻⁷ * L) / 0.1 = 120π × 10⁻⁷ * L N/m
Therefore, the magnitude of the force per unit length on the bottom wire is 120π × 10⁻⁷ * L N/m.
Using the right-hand grip rule, the correct direction of the force on the bottom wire is "out of the page."
Remember to substitute the appropriate length value (L) to get the final answers.