Question

A metal rod of length $l$ moves with a constant speed $v = 2$ m/s parallel to a long wire. Current in the long wire is 5 A. Induced emf a cross ends of the rod is ln($P$) microvolts. Find value of $P$.

Answer 1

16

Answer 2

The magnetic field at a distance $r$ from the wire is $B(r) = \frac{\mu_0 i}{2\pi r}$. The rod moves perpendicular to this field. The induced emf is given by $\mathcal{E} = \int_{r_1}^{r_2} v B(r) dr$. The distances from the wire to the rod's ends are $r_1 = l/3$ and $r_2 = l/3 + l = 4l/3$. Thus, $\mathcal{E} = \int_{l/3}^{4l/3} v \frac{\mu_0 i}{2\pi r} dr = \frac{\mu_0 i v}{2\pi} \ln\left(\frac{4l/3}{l/3}\right) = \frac{\mu_0 i v}{2\pi} \ln(4)$. Substituting the given values: $i = 5$ A, $v = 2$ m/s, and $\mu_0 = 4\pi \times 10^{-7}$ T m/A. $\mathcal{E} = \frac{(4\pi \times 10^{-7})(5)(2)}{2\pi} \ln(4) = 20 \times 10^{-7} \ln(4) = 2 \times 10^{-6} \ln(4)$ Volts. This is $2 \ln(4) = \ln(4^2) = \ln(16)$ microvolts. Since the induced emf is $\ln(P)$ microvolts, we have $\ln(P) = \ln(16)$, so $P = 16$.