Faraday's law with Calc 3 integration help

[horsewnoname]

Homework Statement

Suppose an infinitely long wire carrying current ##I=sin_0(\omega t)## is a distance ##a## away from a equilateral triangular circuit with resistance ##R## in the same plane as shown in the figure. Each side of the circuit is length ##b##. I need to find the induced voltage ##V_{ind}## and current ##i## of the triangular circuit.

Homework Equations

Faraday's law ##V_{ind}=-\frac{\mathrm{d} \phi}{\mathrm{d} x}## where magnetic flux ##\phi = \int \boldsymbol{B}\cdot d\boldsymbol{s}## and ##d\boldsymbol{s}## is the surface through which the flux intercepts.

Also relevant is Ohm's law which is simply ##V=IR##.

The Attempt at a Solution

So the first thing I have to do is determine the ##\boldsymbol{B}## produced by the infinite wire. This is simply ##\boldsymbol{B}=\frac{\mu_0 I}{2 \pi x}##.

From this, I can determine the magnetic flux which is the tricky part. To do this, I need to integrate over the area of the triangle which means I need a double integral where the horizontal ##x## limits of integration are given by ##x=a## to ##x=a+\frac{\sqrt3}{2}b## and the vertical ##y## limits of integration are given by ##y=0## to ##y=\frac{-1}{\sqrt3}x+\frac{a}{\sqrt3}+\frac{b}{2}##. I got this equation of a line by splitting the triangle in half as I show in the figure above. The only thing is that I will have to multiply the result by 2 to consider both areas of the triangle, not just half.

It's been a while since I've taken Calc 3, but I eventually got to the result ##\frac{\mu_0I}{\pi} [\frac{-b}{2}+(\frac{a}{\sqrt3}+\frac{b}{2}) ln(\frac{a+\frac{\sqrt3 b}{2}}{a})]##.

From here, I know I can just plug ##\phi## into Faraday's law and get the induced voltage, but I don't know if I did the integral correctly. Is there any way for me to check that I have the correct result?

 

[TSny]

A partial check to see if your expression for y and your limits are correct is to see if you get the correct area for the triangle if you evaluate ∫ydx.

Your explanation of how you set up the integral sounds right. I think your answer for the flux is correct. (Your parentheses look a little out of place.)