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## Homework Statement

I'm currently reviewing some of the old E&M material that I haven't seen in a while and have gotten stuck on this problem. I'm sure I'm just making a simple mistake but I can't seem to make any headway.

Here's the problem: Figure 35-32 shows a copper rod moving with velocity v parallel to a long straight wire carrying a current i. Calculate the induced emf in the rod, assuming v = 5.0m/s, i = 100A, a = 1cm and b = 20cm. Answer: 3.0X10

^{-4}V

## The Attempt at a Solution

The magnetic field of the wire a distance y away is [tex]B = \frac{\mu _{0}i}{2\pi y}[/tex]. Thus the total magnetic field across the length of the rod is [tex]B = \frac{\mu_{0}i}{2\pi}\int_{a}^{b}\frac{dy}{y} = \frac{\mu_{0}i}{2\pi}ln\frac{b}{a}[/tex]. Now the induced emf is [tex]\varepsilon = -\frac{d\Phi _{B}}{dt}=-B\frac{dA}{dt}=-B(b-a)\frac{dx}{dt}=-B(b-a)v[/tex]. Then the total equation is [tex]\varepsilon = \frac{-\mu_{0}iv}{2\pi}(b-a)ln\frac{b}{a}[/tex], the only problem being that this is apparently not the correct answer.