Welcome back — missed you this week. 1. Step 1: Convert the velocity from km/hr to m/s. $$v = 80 \text{ km/hr} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}}$$ $$v = \frac{80000}{3600} \text{ m/s} = \frac{800}{36} \text{ m/s} = \frac{200}{9} \text{ m/s} \approx 22.22 \text{ m/s}$$ Step 2: Convert the magnetic field from microtesla to tesla. $$B = 40 \text{ µT} = 40 \times 10^{-6} \text{ T}$$ Step 3: Calculate the induced EMF using the formula $\mathcal{E} = B L v$. $$\mathcal{E} = (40 \times 10^{-6} \text{ T}) \times (2 \text{ m}) \times \left(\frac{200}{9} \text{ m/s}\right)$$ $$\mathcal{E} = \frac{40 \times 10^{-6} \times 2 \times 200}{9} \text{ V}$$ $$\mathcal{E} = \frac{16000 \times 10^{-6}}{9} \text{ V}$$ $$\mathcal{E} = \frac{16}{9} \times 10^{-3} \text{ V}$$ $$\mathcal{E} \approx 1.777 \times 10^{-3} \text{ V}$$ $$\mathcal{E} \approx 1.78 \text{ mV}$$ The induced EMF is $\boxed{\text{1.78 mV}}$. 2. Step 1: Identify the initial and final magnetic flux. Initial magnetic flux $\Phi_1 = 400 \text{ µWb}$. Since the flux is reversed, the final magnetic flux $\Phi_2 = -400 \text{ µWb}$. Step 2: Calculate the change in magnetic flux $\Delta \Phi$. $$\Delta \Phi = \Phi_2 - \Phi_1 = (-400 \text{ µWb}) - (400 \text{ µWb})$$ $$\Delta \Phi = -800 \text{ µWb} = -800 \times 10^{-6} \text{ Wb}$$ Step 3: Apply Faraday's Law of Induction, $\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}$. Given $N = 1200$ turns and $\Delta t = 0.1 \text{ s}$. $$\mathcal{E} = -(1200) \times \frac{-800 \times 10^{-6} \text{ Wb}}{0.1 \text{ s}}$$ $$\mathcal{E} = (1200) \times \frac{800 \times 10^{-6}}{0.1} \text{ V}$$ $$\mathcal{E} = 1200 \times 8000 \times 10^{-6} \text{ V}$$ $$\mathcal{E} = 9600000 \times 10^{-6} \text{ V}$$ $$\mathcal{E} = 9.6 \text{ V}$$ The induced EMF is $\boxed{\text{9.6 V}}$. Send me the next one 📸