Magnetic Field at a point due to two parallel wires

[smithisize]

Homework Statement

Two long parallel wires are a center-to-center distance of 4.90 cm apart and carry equal anti-parallel currents of 3.70 A. Find the magnetic field intensity at the point P which is equidistant from the wires. (R = 10.00 cm).

http://imageshack.us/a/img11/3812/twoparallelwires.jpg

Homework Equations

dB = μ_o*I*(dL$\vec{}$xdR$\vec{}$)/(4*pi*r^3)

The Attempt at a Solution

Well first I tried multiplying the equation for magnetic field of an infinite line (μ_o*I/(2*pi*r) by two since there are two wires. then I realized that since the current is flowing in opposite directions, the y-component of the field, so to speak, would cancel out, and now I'm stuck.

Here is the solution, but I want to know how to arrive here (and more specifically I'd like to know why we're multiplying the infinite line equation by the y-component): d*I*μ_o/(2*pi*(R^2+ (d/2)^2)

 

[tiny-tim]

hi smithisize!

… then I realized that since the current is flowing in opposite directions, the y-component of the field, so to speak, would cancel out …

no, if you draw a vector diagram (with arrows), you'll find the x components cancel out

 

[smithisize]

hi smithisize! no, if you draw a vector diagram (with arrows), you'll find the x components cancel out

Well, the vector equations for dr are R$\hat{i}$ -(d/2)$\hat{j}$ and R$\hat{i}$+(d/2)$\hat{j}$ therefore the vector sum states that the vertical component cancels out.

But, I ended up figuring it out. If you draw the b-field for the top wire, and manipulate theta (of the upper left corner of a triangle drawn on the top wire) a bit, you can end up with the proper configuration and see that where it all comes from.

 

[tiny-tim]

Well, the vector equations for dr are R$\hat{i}$ -(d/2)$\hat{j}$ and R$\hat{i}$+(d/2)$\hat{j}$

no, if you draw the arrows, you'll see that one goes to the left, and the other to the right,

so it's -R$\hat{i}$ +(d/2)$\hat{j}$ and R$\hat{i}$+(d/2)$\hat{j}$

 

[Nickie]

^(3/2))The magnetic field at a point due to two parallel wires is a classic example of applying the superposition principle in electromagnetism. The principle states that the total field at a point is equal to the vector sum of the individual fields produced by each source.

In this case, we have two wires carrying equal anti-parallel currents, which means that the magnetic field produced by one wire will be in the opposite direction to the field produced by the other wire. This leads to a cancellation of the y-component of the field, as you correctly pointed out.

To arrive at the given solution, we need to consider the contribution of each wire separately. Let's first consider the wire on the left (let's call it wire A). The magnetic field produced by wire A at point P can be calculated using the equation for the magnetic field of an infinite line:

dB_A = μo*I_A/(2*pi*r)

Where μo is the permeability of free space, I_A is the current in wire A, and r is the distance from wire A to point P. However, we need to take into account the fact that the point P is not directly above or below wire A, but is at a distance d/2 to the side. This means that the actual distance from wire A to point P is given by the Pythagorean theorem as:

d_A = √(r^2 + (d/2)^2)

Substituting this into the equation for dB_A, we get:

dB_A = μo*I_A/(2*pi*(r^2 + (d/2)^2))

Similarly, we can calculate the magnetic field produced by the other wire (let's call it wire B) as:

dB_B = μo*I_B/(2*pi*(r^2 + (d/2)^2))

Where I_B is the current in wire B. Since the currents in wire A and wire B are equal and opposite, we can combine these two equations to get the total magnetic field at point P as:

dB_total = dB_A + dB_B = μo*I/(2*pi*(r^2 + (d/2)^2))

Where I = I_A = -I_B is the total current carried by both wires. This is the same equation as the one given in the solution, but with a negative sign for the current to account for the opposite direction of the currents in the two wires.

In summary, to