Question #cc69d

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Answer 1 · 2018-02-05T08:56:55

$9.44 \cdot 10^{- 3} W b l s^{- 1}$ $9.44*10^-3Wbcolor(white)(l)s^-1$

From Faraday's law, we know that:
$\frac{d N ϕ}{d t} \propto V$ $(dNphi)/(dt)proptoV$

We also know that $V = I R$ $V=IR$

Given resistivity, we also know that:
$R = \frac{ρ l}{A}$ $R=(rhol)/A$

Combing these gives us:
$\frac{d N ϕ}{d t} \propto \frac{I ρ l}{A}$ $(dNphi)/(dt)propto(Irhol)/A$

$rho=1.97*10^-8Omegam$
$I=8.9A$

$l=2pir=pi*9.1/100=(91pi)/1000m$
$A=pir^2=pi(2.6/2000)^2=pi(1.69*10^-6)m^2$

$(dNphi)/(dt)propto(8.9(1.97*10^-8)((91pi)/1000))/(pi(1.69*10^-6))=9.44*10^-3Wbcolor(white)(l)s^-1$

The magnetic field must change at a rate of $9.44*10^-3$ Webers per second.