Question

In figure a wire perpendicular to a long straight wire is moving parallel to it with a speed $v=10\text{m/s}$ in the direction of the current flowing in the wire. The current is $10\text{A}$. What is the magnitude of the potential difference (in $\mu \text{V}$) between the ends of the moving wire?
Take $\ln 2=0.7$

Answer 1

Let us assume a small element $dx$ at a distance $x$ from the current carrying wire.
$v=10\text{m/s}$
$I=10\text{A}$

Now potential difference across small elements is
$dV=Bvdx$ $\{\because e=Blv\}$
$\because$ magnetic field for long wire at a point $x$ away from wire is
$B=\frac{{\mu }_{0}i}{2\pi d}=\frac{{\mu }_{0}i}{2\pi x}$
$\therefore dV=\frac{{\mu }_{0}Iv}{2\pi x}.dx$
Now the p.d. across the moving wire is
${\int }_0^{V}dV={\int }_{r_1}^{r_2}Bdxv={\int }_{r_1}^{r_2}\frac{{\mu }_{0}i}{2\pi x}vdx$
$V=\frac{{\mu }_{0}iv}{2\pi }\ln \left(\frac{10}{5}\right)$
$V=\frac{4\pi \times {10}^{-7}\times 10\times 10\ln 2}{2\pi }=2\times {10}^{-5}ln2$
$V=1.4\times {10}^{-5}V=14\mu \text{V}$