Two long, parallel wires carry current in the x-y plane. One wire carries 30 A to the left along the x-axis. The other carries 50 A to the right along a parallel line at y = 0.28 m. At what y-axis position in meters is the magnetic field equal to zero?
I saw that how Elena doing, but when I tried it was wrong. She found that 0.105. At first, I was also tried in same way. It was wrong.. any other information pls
y= -0.42
The region isn't between the wires
To find the y-axis position where the magnetic field is equal to zero, we can use the equation for the magnetic field produced by a current-carrying wire:
B = (μ0 * I) / (2πr)
where B is the magnetic field, μ0 is the permeability of free space (4π × 10^-7 T·m/A), I is the current, and r is the distance from the wire.
Let's denote the distance from the wire carrying 30 A to be x and the distance from the wire carrying 50 A to be y. We want to find the value of y where the magnetic field is zero.
Using the equation for the magnetic field, we can set up the following equation:
(μ0 * 30 A) / (2πx) = (μ0 * 50 A) / (2πy)
Simplifying this equation, we can cancel out the common terms:
30 / x = 50 / y
Now we can solve for y:
y = (50 * x) / 30
Substituting the given x value for the wire carrying 30 A:
y = (50 * 0.28 m) / 30
Calculating this expression, we find:
y ≈ 0.467 m
Therefore, the y-axis position where the magnetic field is equal to zero is approximately 0.467 meters.
To find the y-axis position where the magnetic field is equal to zero in this scenario, we need to apply Ampere's Law. Ampere's Law states that the magnetic field around a closed loop is proportional to the current passing through the loop.
We can consider a circular Amperian loop centered at the origin (0,0) with a radius of r. By convention, the current enclosed by the loop is positive if it is flowing in a counterclockwise direction and negative if flowing clockwise.
Following Elena's approach, let's consider the magnetic field generated by the 30 A current-carrying wire on the Amperian loop. According to the right-hand rule, the magnetic field lines created by the current-carrying wire will circulate counterclockwise around the wire when viewing it from above.
Now, let's apply Ampere's Law for this particular Amperian loop:
∮ B · dl = μ₀ * I_enclosed,
where B is the magnetic field vector, dl is the differential path length around the loop, μ₀ is the vacuum permeability constant (4π × 10^(-7) T·m/A), and I_enclosed is the current enclosed by the loop.
Since we only have one current-carrying wire inside the loop, I_enclosed will be equal to the current flowing in that wire, which is 30 A. Therefore, we can write:
∮ B · dl = μ₀ * 30 A.
Now let's consider the magnetic field generated by the 50 A current-carrying wire. Using the right-hand rule, we can determine that the magnetic field lines from this wire will circulate clockwise around it when viewing it from above.
Applying Ampere's Law again for the same Amperian loop:
∮ B · dl = μ₀ * (-50 A),
where -50 A indicates that the current is flowing clockwise and in the opposite direction to the previous wire. Note that in this equation, the negative sign is taken into account to reflect the direction of the current.
For the magnetic field to be zero at a certain y-axis position, the total magnetic field due to both wires at that position must be zero. It means the magnitudes of the magnetic fields due to each wire must be equal.
Let's integrate the magnetic field around this Amperian loop for both wires:
∮ B₁ · dl + ∮ B₂ · dl = μ₀ * I₁ + μ₀ * I₂,
where B₁ is the magnetic field due to the 30 A current-carrying wire, B₂ is the magnetic field due to the 50 A current-carrying wire, I₁ is the current flowing in the 30 A wire, and I₂ is the current flowing in the 50 A wire.
Simplifying this expression, we get:
∮ (B₁ + B₂) · dl = μ₀ * (I₁ - I₂).
Since we want the magnetic field to be zero at the y-axis position, we have:
∮ (B₁ + B₂) · dl = 0.
Therefore, μ₀ * (I₁ - I₂) must also be zero:
μ₀ * (I₁ - I₂) = 0.
Now we can calculate the y-axis position where the magnetic field is equal to zero.
Given that I₁ = 30 A and I₂ = -50 A, plugging these values into the equation:
μ₀ * (30 A - (-50 A)) = 0.
Simplifying further, we get:
μ₀ * (30 A + 50 A) = 0.
Since μ₀ is a non-zero constant, we can conclude that (30 A + 50 A) must be equal to zero:
30 A + 50 A = 0.
However, since the sum of 30 A and 50 A is equal to 80 A, not zero, it means that there is no y-axis position at which the magnetic field is equal to zero in this scenario.