32: Magnetic Field from Two Wires

Figure 1

Figure 2

INTRO:
From the Biot-Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by
B_wire = μ₀I / 2πd ,

where μ0 (=4π×10⁻⁷ T⋅m/A) is the permeability constant, I is the current in the wire, and d is the distance from the wire to the location at which the magnitude of the magnetic field is being calculated.

The same result can be obtained from Ampere's law as well.

The direction of vector B⃗ can be found using the right-hand rule.

In this problem, you will be asked to calculate the magnetic field due to a set of two wires with antiparallel currents as shown in the diagram (figure 1). Each of the wires carries a current of magnitude I. The current in wire 1 is directed out of the page and that in wire 2 is directed into the page. The distance between the wires is 2d. The x-axis is perpendicular to the line connecting the wires and is equidistant from the wires.

As you answer the questions posed here, try to look for a pattern in your answers.

PART A:
Which of the vectors best represents the direction of the magnetic field created at point K (figure 1) by wire 1 alone?
Enter the number of the vector with the appropriate direction (figure 2).

SOLUTION:
For this, begin by pointing your thumb towards yourself since the current in wire 1 is directed out of the page. This means that the magnetic field will follow the direction of your fingers, counterclockwise. At point K, the question asks, what is the direction of the magnetic field created by only wire 1? Wouldn't that just be the tangential direction of the circular magnetic field at that point? Yes, it would be pointing to the right, or to choose one of the given vectors, vector 3.

PART B:
Which of the vectors best represents the direction of the magnetic field created at point K (figure 1) by wire 2 alone?
Enter the number of the vector with the appropriate direction (figure 2).

SOLUTION:
Do the exact same thing you did for the last problem except for this time point your thumb into the page, resulting in your fingers curling clockwise. At point K, the tangent of this circle would be pointing to the right. This means that vector 3 is our answer again. 

PART C:
Which of these vectors best represents the direction of the net magnetic field created at point K (figure 1) by both wires?
Enter the number of the vector with the appropriate direction (figure 2).

SOLUTION:
Think about the two previous answers. Since both of them were pointing to the right, vector 3, the only logical answer is that the net magnetic field is also pointing in the direction of vector 3.

PART D:
Find the magnitude of the magnetic field B_1K created at point K by wire 1.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION: 
Since all of the values are already in the formula, B_wire = μ₀I / 2πd, we don't have to do anything else. The answer is μ₀I / 2πd
keep in mind that since K is equidistant from both wires, B_1K = B_2K

PART E: 
Find the magnitude of the net magnetic field B_K created at point K by both wires.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION:
B_K = B_1K + B_2K
[μ₀I / 2πd] + [μ₀I / 2πd] = μ₀I / πd

PART F:
Point L is located a distance d(√2) from the midpoint between the two wires. Find the magnitude of the magnetic field B_1L created at point L by wire 1.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION: 
The first thing we must do in this problem is determine the distance between point L and either one of the wires (equidistant again). We do this by setting up a right triangle and solving for the hypotenuse. Using Pythagorean's theorem, d²+[d(√2)]² = r²
r=√(d²+2d²) = √(3d²) = d(√3)
the magnetic field at this point L would be μ₀I / [2πd(√3)]

PART G:
Point L is located a distance d(√2) from the midpoint between the two wires. Find the magnitude of the magnetic field B_L created at point L by both wires.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION: 
We can use the value from the previous part. Do not forget that this value is a vector, so we must take the x and y components into consideration if we are to determine the correct net magnetic field. We can set up a proportional right triangle using the ratio of the distances. The hypotenuse would be √3, the adjacent component would be 1 and the opposite component would be √2
y_1L = B_1L⋅sin(θ) = μ₀I⋅ 1/[2πd(√3)]⋅[√2/√3]= μ₀I⋅(√2)/ [6πd]
x_1L = B_1L⋅cos(θ) = μ₀I⋅ 1/[2πd(√3)]⋅1/(√3) = μ₀I/[6πd]
y_2L = B_2L⋅sin(-θ) = μ₀I⋅ 1/[2πd(√3)]⋅[√2/√3]= -μ₀I⋅(√2)/ [6πd]
x_2L = B_2L⋅cos(-θ) = μ₀I⋅ 1/[2πd(√3)]⋅1/(√3) = μ₀I/ [6πd]
Now that we have the x and y components, we are able to determine the magnetic field at point L
B_x = x_1L + x_2L = μ₀I/[6πd] + μ₀I/[6πd] = μ₀I/[3πd]
B_y = y_1L + y_2L = μ₀I⋅(√2)/ [6πd] + -μ₀I⋅(√2)/ [6πd] = 0

This leaves us with the magnitude of the vector <μ₀I/[3πd], 0>
which is μ₀I/[3πd]