# Question #cc69d

##### 1 Answer
Feb 5, 2018

$9.44 \cdot {10}^{-} 3 W b \textcolor{w h i t e}{l} {s}^{-} 1$

#### Explanation:

From Faraday's law, we know that:
$\frac{\mathrm{dN} \phi}{\mathrm{dt}} \propto V$

We also know that $V = I R$

Given resistivity, we also know that:
$R = \frac{\rho l}{A}$

Combing these gives us:
$\frac{\mathrm{dN} \phi}{\mathrm{dt}} \propto \frac{I \rho l}{A}$

$\rho = 1.97 \cdot {10}^{-} 8 \Omega m$
$I = 8.9 A$

$l = 2 \pi r = \pi \cdot \frac{9.1}{100} = \frac{91 \pi}{1000} m$
$A = \pi {r}^{2} = \pi {\left(\frac{2.6}{2000}\right)}^{2} = \pi \left(1.69 \cdot {10}^{-} 6\right) {m}^{2}$

$\frac{\mathrm{dN} \phi}{\mathrm{dt}} \propto \frac{8.9 \left(1.97 \cdot {10}^{-} 8\right) \left(\frac{91 \pi}{1000}\right)}{\pi \left(1.69 \cdot {10}^{-} 6\right)} = 9.44 \cdot {10}^{-} 3 W b \textcolor{w h i t e}{l} {s}^{-} 1$

The magnetic field must change at a rate of $9.44 \cdot {10}^{-} 3$ Webers per second.