Let's solve the problem in **2 steps** as requested. --- ## **Step 1: Find the Induced EMF (\(\mathcal{E}\))** **Explanation:** The change in magnetic field induces an emf in the loop, given by Faraday’s law: \[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -A\frac{dB}{dt} \] where: - \(A = 11 \times 10^{-4}\) m² (area of the loop) - \(\frac{dB}{dt}\) is the rate of change of the magnetic field. From the given graph (b): - \(B\) increases linearly from to \(B_s = 10\) mT in \(t_s = 6.1\) s. - Thus, \(\frac{dB}{dt} = \frac{10 \times 10^{-3}}{6.1}\) T/s. Now, plug in the values: \[ \mathcal{E} = A \left(\frac{dB}{dt}\right) = 11 \times 10^{-4} \cdot \frac{10 \times 10^{-3}}{6.1} \] \[ \mathcal{E} = 11 \times 10^{-4} \cdot \frac{10}{6.1} \times 10^{-3} \] \[ \mathcal{E} = 11 \times 10^{-4} \cdot 1.639 \times 10^{-3} \] \[ \mathcal{E} \approx 1.8 \times 10^{-6} \text{ V} \] --- ## **Step 2: Find the Resistance Using the Total Charge Passed** **Explanation:** The total charge \(q\) passing through the loop in time \(t_s\) is given in the graph (c) as \(q_s = 6.7\) mC \(= 6.7 \times 10^{-3}\) C. We use the relation: \[ q = \frac{\mathcal{E} t_s}{R} \] So, \[ R = \frac{\mathcal{E} t_s}{q} \] Plug in the values: - \(\mathcal{E} \approx 1.8 \times 10^{-6}\) V (from Step 1) - \(t_s = 6.1\) s - \(q = 6.7 \times 10^{-3}\) C \[ R = \frac{1.8 \times 10^{-6} \times 6.1}{6.7 \times 10^{-3}} \] \[ R = \frac{1.098 \times 10^{-5}}{6.7 \times 10^{-3}} \] \[ R \approx 1.64 \times 10^{-3} \, \Omega \] --- ## **Final Answer** **The loop's resistance is:** \[ \boxed{1.6 \times 10^{-3}~\Omega} \]