Answer
$\mathscr{E} = 2.40\times 10^{-4}~V$
Work Step by Step
We can write an expression for the magnetic field around a wire with a current:
$B = \frac{\mu_0~I}{2\pi~r}$
We can find an expression for the magnetic flux when the width of the loop is $w$:
$\Phi = \int~B\cdot dA$
$\Phi = \int_{a}^{a+L}~\frac{\mu_0~I}{2\pi~r}~w~dr$
$\Phi = \frac{\mu_0~I}{2\pi}~w~ln (r) \Big \vert_{a}^{a+L}$
$\Phi = \frac{\mu_0~I}{2\pi}~w~ln (r) \Big \vert_{0.010}^{0.11}$
$\Phi = \frac{\mu_0~I~w}{2\pi}~[ln (0.11)-ln(0.010)]$
$\Phi = \frac{\mu_0~I~w}{2\pi}~(2.398)$
We can find the magnitude of the induced emf:
$\mathscr{E} = \frac{d\Phi}{dt}$
$\mathscr{E} = (2.398)~\frac{\mu_0~I}{2\pi}~\frac{dw}{dt}$
$\mathscr{E} = (2.398)~\frac{\mu_0~I~v}{2\pi}$
$\mathscr{E} = (2.398)~\frac{(4\pi\times 10^{-7}~H/m)~(100~A)~(5.00~m/s)}{2\pi}$
$\mathscr{E} = 2.40\times 10^{-4}~V$
