Let’s solve **Problem 7: Power Grid Failure** step by step: --- ### **(a) Find the current as a function of time from the magnetic flux** #### **Step 1: Magnetic Flux Through the Loop** - Area of loop: \(A = w \ell\) - \(N\) loops: Total area = \(N A\) - Uniform magnetic field: \(\vec{B}\) - Angle between normal to area and \(\vec{B}\): \(\theta\) \[ \Phi_B = N B A \cos\theta \] As the loop rotates with angular speed \(\omega\): \[ \theta = \omega t \implies \Phi_B(t) = N B A \cos(\omega t) \] #### **Step 2: Induced EMF (Faraday’s Law)** \[ \mathcal{E}_{ind} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt} [N B A \cos(\omega t)] = N B A \omega \sin(\omega t) \] #### **Step 3: Current Produced** Ohm’s law: \(I_{ind} = \frac{\mathcal{E}_{ind}}{R}\) \[ I_{ind}(t) = \frac{N B A \omega}{R} \sin(\omega t) \] **Final answer (a):** \[ \boxed{I_{ind}(t) = \frac{N B w \ell \omega}{R} \sin(\omega t)} \] --- ### **(b) Net Torque on the Coil as a Function of Time** - Magnetic moment: \(\vec{\mu} = N I \vec{A}\) - Magnitude: \(\mu = N I A\) - Torque: \(\vec{\tau} = \vec{\mu} \times \vec{B}\) - Magnitude: \(\tau = \mu B \sin\theta\) Plug in \(I_{ind}(t)\) from (a): \[ \tau_{ind}(t) = N I_{ind}(t) A B \sin(\omega t) \] Substitute \(A = w \ell\) and \(I_{ind}(t)\): \[ \tau_{ind}(t) = N [\frac{N B w \ell \omega}{R} \sin(\omega t)] (w \ell) B \sin(\omega t) = \frac{N^2 B^2 (w \ell)^2 \omega}{R} \sin^2(\omega t) \] **Final answer (b):** \[ \boxed{\tau_{ind}(t) = \frac{N^2 B^2 w^2 \ell^2 \omega}{R} \sin^2(\omega t)} \] --- ### **(c) What happens as more buildings increase their load (parallel resistors)?** **Explanation:** - More buildings = More parallel resistors = **total resistance decreases**. - For a fixed voltage (emf), **current drawn from the generator increases** as load increases. --- ### **(d) How does the generator’s rotation rate change as load increases?** **Explanation:** - As load increases, current increases, so the **back-torque** (from part (b)) increases. - This torque opposes the rotation, so **the generator slows down** unless more mechanical input is provided. --- ### **(e) Using Lenz’s Law, what is the coil’s response as load increases?** **Explanation:** - By **Lenz’s Law**, the induced current produces a torque opposing the motion. - As load increases: - Induced EMF stays the same (if \(\omega\) is fixed), but current increases as \(R\) decreases. - **Induced current increases** (\(I_{ind}\) increases). - **Induced torque (\(\tau_{ind}\)) increases** (from part (b)). - This increased torque opposes rotation more, so **rotation rate decreases** unless additional mechanical power is supplied. --- ## **Summary Table** | Part | Answer | |------|--------| | (a) | \(I_{ind}(t) = \frac{N B w \ell \omega}{R} \sin(\omega t)\) | | (b) | \(\tau_{ind}(t) = \frac{N^2 B^2 w^2 \ell^2 \omega}{R} \sin^2(\omega t)\) | | (c) | Current increases as load increases (total resistance drops) | | (d) | Rotation rate decreases as load increases (more opposing torque) | | (e) | Induced current and torque increase as load increases (Lenz’s Law), causing the generator to slow down unless more input power is provided |