Question

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; its slow sinusoidal variation in time is 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.

Short answer

Answer: The induced potential difference across the inner loop is given by: $$V_{2}(t) = \frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} r}$$ The current flowing through the inner loop is given by: $$i(t) = \frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} R_{2} r}$$

Step-by-step solution

Find the magnetic field at the center of the outer loop

To find the potential difference induced in the inner loop, we first need to determine the magnetic field (\(\textbf{B}\)) at the center of the outer conducting loop. To do this, we can use Ampere's law, which relates the circulation of the magnetic field around a closed loop to the current passing through it: $$\oint_{\textbf{C}} \textbf{B} \cdot d\textbf{l} = \mu_{0}I_{\text{enclosed}}$$ Since we are given that \(V(t) = V_{0} \sin(\omega t)\) is the voltage applied to the larger loop, we can use Ohm's Law to find the current passing through this outer loop: $$I_{1}(t) = \frac{V(t)}{R_{1}} = \frac{V_{0} \sin(\omega t)}{R_{1}}$$ Now, we know that for concentric circular loops, the magnetic field is radial and symmetrical around the center. Therefore, we can consider the circulation of the magnetic field around a circle of radius \(r\) that lies between \(a\) and \(b\). For this situation, we have: $$B(r) 2\pi r = \mu_{0} I_{1} = \mu_{0} \frac{V_{0} \sin(\omega t)}{R_{1}}$$ Solving for the magnetic field B(r) at the center: $$B(r) = \frac{\mu_{0} V_{0} \sin(\omega t)}{2\pi R_{1} r}$$

Calculate the magnetic flux through the inner loop

To find the potential difference in the inner loop, we need to find the magnetic flux \(\Phi_{\text{magnetic}}(t)\) through it. Since we are given that the magnetic field throughout the inner loop is uniform and equal to the field at the center of the loop, the magnetic flux through the inner loop is: $$\Phi_{\text{magnetic}}(t) = B(r) \pi a^2 = \frac{\mu_{0} V_{0} \sin(\omega t)}{2\pi R_{1} r} \pi a^2$$

Find the induced potential difference in the inner loop

Now, we can use Faraday's Law of electromagnetic induction to find the induced potential difference across the inner loop: $$V_{2}(t) = -\frac{d\Phi_{\text{magnetic}}}{dt}$$ Taking the time derivative of the magnetic flux through the inner loop: $$\frac{d\Phi_{\text{magnetic}}}{dt} = -\frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} r}$$ This gives us the induced potential difference across the inner loop: $$V_{2}(t) = \frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} r}$$

Calculate the current flowing through the inner loop

Finally, we can use Ohm's Law to find the current flowing through the inner loop: $$i(t) = \frac{V_{2}(t)}{R_{2}} = \frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} R_{2} r}$$ Thus, we have found expressions for both the induced potential difference across the inner loop: $$V_{2}(t) = \frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} r}$$ and the current flowing through it: $$i(t) = \frac{\mu_{0} V_{0} a^2 \omega \cos(\omega t)}{2 R_{1} R_{2} r}$$