A circular conducting loop with radius \(a\) and resistance \(R_{2}\) is concentric with a circular conducting loop with radius \(b \gg a(b\) much greater than \(a\) ) and resistance \(R_{1}\). A time-dependent voltage is applied to the larger loop, having a slow sinusoidal variation in time given by \(V(t)=V_{0} \sin \omega t\) where \(V_{0}\) and \(\omega\) are constants with dimensions of voltage and inverse time, respectively. Assuming that the magnetic field throughout the inner loop is uniform (constant in space) and equal to the field at the center of the loop, derive expressions for the potential difference induced in the inner loop and the current \(i\) through that loop.

The current \(I_{1}\) in the outer loop can be determined using Ohm's law. Ohm's law states that the voltage across a resistor is equal to the product of the resistance and the current through it. Thus, we have: $$I_{1}(t) = \frac{V(t)}{R_{1}} = \frac{V_{0} \sin \omega t}{R_{1}}$$

The magnetic field produced by a circular current loop at its center can be obtained by applying Biot-Savart law or Ampere's law. Here we will use Ampere's law, which states that the integral of the magnetic field around a closed-loop path is equal to the total enclosed current multiplied by the permeability of free space, \(\mu_{0}\). For a circular loop with current \(I_{1}\) and radius \(b\), the magnetic field at the center is given by: $$B(t) = \mu_{0}\frac{I_{1}(t)}{2b} = \frac{\mu_{0} V_{0}}{2b R_{1}} \sin \omega t$$

Faraday's law of electromagnetic induction states that the induced electromotive force (EMF) in a closed loop is equal to the rate of change of the magnetic flux through the loop. For the inner circular loop with radius \(a\), the potential difference induced across it can be given by: $$V_{induced}(t) = - \frac{d\Phi}{dt}$$ where \(\Phi\) is the magnetic flux through the inner loop and the negative sign indicates Lenz's law. We know that the magnetic field throughout the inner loop is assumed to be uniform and equal to the field at the center of the loop. Therefore, the magnetic flux through the inner loop is: $$\Phi(t) = B(t)\pi a^2$$ Now, take the derivative of \(\Phi(t)\) with respect to \(t\): $$\frac{d\Phi}{dt} = \pi a^2 \frac{d}{dt}\left(\frac{\mu_{0} V_{0}}{2b R_{1}} \sin \omega t\right)$$ By applying the chain rule, we get: $$\frac{d\Phi}{dt} = \pi a^2 \left(\frac{\mu_{0} V_{0}}{2b R_{1}}\right)\omega\cos\omega t$$ Therefore, the induced potential difference in the inner loop is given by: $$V_{induced}(t) = - \pi a^2 \left(\frac{\mu_{0} V_{0}}{2b R_{1}}\right)\omega\cos\omega t$$

To find the current \(i\) through the inner loop, we can use Ohm's law, which states that the voltage across the resistor is equal to the product of the resistance and the current through it. Thus, we have: $$i(t) = \frac{V_{induced}(t)}{R_{2}} = -\pi a^2 \left(\frac{\mu_{0} V_{0}}{2b R_{1}R_{2}}\right)\omega\cos\omega t$$ The final expressions for the potential difference induced in the inner loop and the current through that loop are: $$V_{induced}(t) = - \pi a^2 \left(\frac{\mu_{0} V_{0}}{2b R_{1}}\right)\omega\cos\omega t$$ $$i(t) = -\pi a^2 \left(\frac{\mu_{0} V_{0}}{2b R_{1}R_{2}}\right)\omega\cos\omega t$$