Let's break down the problem step by step. --- ## **Step 1: the Scenario** - A loop of material is wrapped around a patient's chest. - The loop is placed in a constant magnetic field (\( B \)), and as the patient inhales, the loop's area (\( A \)) changes. - The device detects changes in the magnetic flux (\( \Phi = BA \)) as an induced emf (\( \mathcal{E} \)). --- ## **Step 2: Relevant Equation - Faraday's Law** The induced emf in a loop is: \[ \mathcal{E} = -N \frac{d\Phi}{dt} \] Where: - \( N \) = number of loops, - \( \Phi = BA \) (if \( B \) is constant and perpendicular to the loop), - \( \frac{d\Phi}{dt} = B \frac{dA}{dt} \) (since \( B \) is constant). So, \[ \mathcal{E} = -N B \frac{dA}{dt} \] --- ## **Step 3: Problem 1** **If the minimum emf that can be reliably detected is \( 2.00 \times 10^{-6} \) V, what is the magnitude \( B_{\text{min}} \) of the minimum external magnetic field needed (assuming \( N=1 \))?** Given: - \( \mathcal{E}_{\text{min}} = 2.00 \times 10^{-6} \) V - \( \Delta A = 5.00 \, \text{cm}^2 = 5.00 \times 10^{-4} \, \text{m}^2 \) - \( \Delta t = 1.50 \) s - \( N = 1 \) Calculate \( \frac{dA}{dt} \): \[ \frac{dA}{dt} = \frac{\Delta A}{\Delta t} = \frac{5.00 \times 10^{-4}}{1.50} = 3.33 \times 10^{-4} \, \text{m}^2/\text{s} \] Solve for \( B_{\text{min}} \): \[ \mathcal{E}_{\text{min}} = N B_{\text{min}} \frac{dA}{dt} \] \[ B_{\text{min}} = \frac{\mathcal{E}_{\text{min}}}{N \frac{dA}{dt}} \] Plug in values: \[ B_{\text{min}} = \frac{2.00 \times 10^{-6}}{1 \times 3.33 \times 10^{-4}} \] \[ B_{\text{min}} = 6.00 \times 10^{-3} \, \text{T} \] **Final Answer for (1):** \[ \boxed{6.00 \times 10^{-3} \ \text{T}} \] --- ## **Step 4: Problem 2** **The solenoid can produce at most \( 3.50 \times 10^{-3} \) T. What is the minimum number \( N_{\text{min}} \) of wire loops that will allow the device to function as desired?** Given: - \( B = 3.50 \times 10^{-3} \) T - All other values same as before. \[ \mathcal{E}_{\text{min}} = N_{\text{min}} \cdot B \cdot \frac{dA}{dt} \] Solve for \( N_{\text{min}} \): \[ N_{\text{min}} = \frac{\mathcal{E}_{\text{min}}}{B \frac{dA}{dt}} \] Plug in values: \[ N_{\text{min}} = \frac{2.00 \times 10^{-6}}{3.50 \times 10^{-3} \times 3.33 \times 10^{-4}} \] \[ = \frac{2.00 \times 10^{-6}}{1.166 \times 10^{-6}} \] \[ = 1.715 \] Since you can't have a fraction of a loop, **round up to the next whole number**: \[ N_{\text{min}} = 2 \] **Final Answer for (2):** \[ \boxed{2} \] --- ## **Summary Table** | Part | Answer | |------|--------| | 1 | \( 6.00 \times 10^{-3} \) T | | 2 | 2 loops |