# Homework Help: Magnetic flux between 2 parallel wires

Tags:
1. Apr 10, 2017

### lowea001

1. The problem statement, all variables and given/known data
Consider two long parallel wires each of radius a with a separation distance d between them. They carry current I in opposite directions. Calculate the magnetic flux through a section of length l, ignoring magnetic field inside the wires.

My confusion lies in trying to reconcile two different approaches to solving this problem. The solution provided takes the following approach: calculate the flux due to one wire in this area and then just double it i.e. take the line integral from a to d - a of the magnetic field and multiply by l (and then double this value). My own approach was to calculate the strength of the magnetic field due to both wires at a distance r, take the line integral from a to d/2, multiply the value by l (and then double this value). However, I cannot get the same solution by doing this.

Their solution: $$\Phi_b = \frac{\mu_o l I}{\pi}ln\left(\frac{d-a}{a}\right)$$
My solution: $$\Phi_b = \frac{\mu_o lI}{\pi}ln\left(\frac{d^2}{4a(d-a)}\right)$$

2. Relevant equations
For a long wire carrying current I, $B = \frac{\mu_o I}{2\pi r}$

3. The attempt at a solution
Their solution: $$\Phi_b = 2 \times \frac{\mu_o lI}{2\pi}\int_{a}^{d-a}\frac{1}{r}dr = \frac{\mu_o lI}{\pi}ln\left(\frac{d-a}{a}\right)$$
My solution: $$\Phi_b = 2 \times \frac{\mu_o lI}{2\pi}\int_{a}^{d/2}\left(\frac{1}{r}+\frac{1}{d-r}\right)dr = \frac{\mu_o lI}{\pi}ln\left(\frac{d^2}{4a(d-a)}\right)$$

I apologize in advance if this is a trivial maths error on my part, but I'm hoping that I am correct in my intuition that these two approaches should give the same result.

2. Apr 10, 2017

### lowea001

Haha five seconds after posting this I found my own mistake but I have decided to leave this up here as a demonstration of how the approaches eventually correspond to the same answer: my mistake was in evaluating the integral $\int_{a}^{d/2}\left(\frac{1}{r}+\frac{1}{d-r}\right)dr = ln\left(\frac{d}{2a}\right) + \left(-ln\left(\frac{d}{2(d-a)}\right)\right)=ln\left(\frac{d-a}{a}\right)$, as expected. I missed the negative sign in evaluating the integral of $\frac{1}{d-r}$, whoops.