Answer
$6.9 \times 10^{-5}$ Amps
Work Step by Step
Recall that Faraday's law is
$\displaystyle \varepsilon_{induced} = |\frac{d\Phi_B}{dt}| = |\vec{A}\frac{d\vec{B}}{dt} + \vec{B}\frac{d\vec{A}}{dt}|$
In this problem, the magnetic field is uniform (and therefore constant) and we know that the volume is changing so the cross-sectional area must be changing too.
$\displaystyle \varepsilon_{induced} = |\vec{B}\frac{d\vec{A}}{dt}|$
Since we know the total resistance, we can say
$I(t) = \displaystyle \frac{\varepsilon_{induced}}{R} = \frac{\vec{B}}{R}\cdot\frac{d\vec{A}}{dt}$
Finding $\displaystyle \frac{dA}{dt}$ is a little tricky, but we can start with the cross-sectional area of the sphere. The cross-sectional area can be represented as
$\displaystyle \pi r^2 = A_{\perp} = \vec{A}$
$\displaystyle \frac{d}{dt}[A_{\perp}] = 2\pi \frac{d}{dt}[r^2]$
$\displaystyle \frac{d\vec{A}}{dt} = 2\pi r \frac{dr}{dt}$
and we also know that the volume of a sphere is
$V_{sphere} = \displaystyle \frac{4}{3}\pi r^3$
$\displaystyle \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
$\therefore \displaystyle \frac{1}{4\pi r^2} \frac{dV}{dt} = \frac{dr}{dt}$
now we can express $\displaystyle \frac{d\vec{A}}{dt}$ in terms of $\displaystyle \frac{dV}{dt}$ by substituting it for $\displaystyle \frac{dr}{dt}$
$\displaystyle \frac{d\vec{A}}{dt} = 2\pi r (\frac{1}{4\pi r^2} \frac{dV}{dt}) = \frac{1}{2r}\frac{dV}{dt}$
So now
$\displaystyle I = \frac{\vec{B}}{R}\cdot\frac{d\vec{A}}{dt} = \frac{\vec{B}}{R}\cdot\frac{1}{2r}\frac{dV}{dt} = \frac{\vec{B}}{2Rr}\frac{dV}{dt}$
You may think that this problem is finished now, but you are wrong. We need to make sure that the radius in our answer is the radius of the spherical balloon AFTER the 2 seconds have
passed. "How do we find this?" you might ask. Let $V(t)$ be the volume of the balloon at any time $t$. Also, what threw me off in doing this question was the conversion between liters
and cubic meters. NOTE: $1$ L = $10^{-3}$ m$^3$
$V(t) = V(0) + \displaystyle\frac{dV}{dt}\cdot t$
What the above equation is saying is that the initial volume, $V(0) = 2.5 $L, of the balloon plus the change in the balloon's volume over a time, $t$, is the overall volume of the balloon.
How does this help us find the radius? By finding the volume of the balloon at $t = 2$, we can use the volume of a sphere to find the radius.
$V(2) = 2.5 + (0.75)(2) = 4$ L = $4\times10^{-3}$ m$^3$
$r |_{t=2} = \displaystyle \sqrt[3] {\frac{3}{4\pi}(4\times10^{-3})} = 0.0985$ m$^3$
Now we can go ahead and plug everything back into our equation. Remember, $\displaystyle \frac{dV}{dt} = 0.75$ L/s $=\displaystyle 0.75\times10^{-3}$ m$^3$/s
$\displaystyle I = \frac{\vec{B}}{2Rr}\frac{dV}{dt} = \frac{45\times10^{-3}}{2(0.0985)(2.5)}(0.75\times10^{-3}) = 6.9\times10^{-5}$ A
