Answer
$5.24C$
Work Step by Step
$\Delta B=-0.670T$
$A_l=(\pi(0.066m)^2)=0.0137m^2$
$\Phi=BA=(-0.670T)(0.0137m^2)$$=-9.16\times10^{-3}Wb$
$A_w=\pi(1.125\times10^{-3})^2=3.976\times10^{-6}m^2$
$l_w=2\pi r=0.415m$
$R=\rho\frac{l}{A}=(1.68\times10^{-8})\frac{0.415m}{3.976\times10^{-6}m^2}=1.75\times10^{-3}\Omega$
$E=\frac{-\Delta\Phi_B}{\Delta t}=\frac{9.16\times10^{-3}Wb}{\Delta t}$
$I=\frac{E}{R}=\frac{\frac{9.16\times10^{-3}Wb}{\Delta t}}{1.75\times10^{-3}\Omega}=\frac{5.24}{\Delta t}$
$Q=I\Delta t=(\frac{5.24}{\Delta t})(1s)=5.24C$
