Two long parallel vertical wires 0.3m apart are. ' placed east-west of one another. The current in the westerly wire is 30A and on the other 20A. The horizontal component of the earths magnetic flux density is 2*10^-5. Find the force per unit length on each wire. (The answers are 0 and 1mN but I don't know why)
I calculated the force on each wire due to the earth. Then I calculated the force due to the wires. Thesr 2 forces are perpendicular so I used pythagoras to find resultant but my answer was wrong
To find the force per unit length on each wire, you need to consider the magnetic field produced by the current in one wire and the magnetic field created by the other wire. The force experienced by a current-carrying wire in a magnetic field is given by the formula:
F = I * L * B * sin(θ)
Where:
F is the force on the wire
I is the current in the wire
L is the length of the wire segment
B is the magnetic field strength
θ is the angle between the direction of the current and the magnetic field
Let's calculate the force on each wire separately:
1. The force on the westerly wire due to the Earth's magnetic field can be calculated as follows:
F_earth = I_west * L * B_earth * sin(90°) (since the angle between the current and Earth's magnetic field is 90°)
= 30A * 1m * (2 * 10^(-5) T) * sin(90°)
= 0
Here, we find that the force on the westerly wire due to the Earth's magnetic field is zero (0).
2. The force on the easterly wire due to the Earth's magnetic field is also zero (0) since the angle between the current and Earth's magnetic field is 90°.
Next, let's calculate the force between the two wires:
The force on the easterly wire due to the westerly wire can be calculated as follows:
F_wire = I_west * L * B_wire * sin(θ)
Where:
B_wire is the magnetic field produced by the westerly wire at the position of the easterly wire.
The magnetic field produced by a long, straight wire at a point perpendicular to the wire is given by Ampere's law:
B_wire = (μ₀ * I_west) / (2π * d)
Where:
μ₀ is the permeability of free space (4π * 10^(-7) T*m/A)
I_west is the current in the westerly wire
d is the separation between the wires (0.3 m in this case)
Let's calculate the force on the easterly wire due to the westerly wire:
F_wire = I_west * L * [(μ₀ * I_west) / (2π * d)] * sin(θ)
The forces on both wires are equal in magnitude but opposite in direction. Therefore, the total force per unit length on each wire (F_total) is:
F_total = F_wire - F_wire
= I_west * L * [(μ₀ * I_west) / (2π * d)] * sin(θ) - I_west * L * [(μ₀ * I_west) / (2π * d)] * sin(θ)
= 0
So, the total force per unit length on each wire due to the interaction between the wires is indeed zero (0).
Hence, the correct answer is 0 mN, not 1 mN.