Let's break the problem into two clear steps--- ### **Step 1: Find the Magnetic Field at the Bar's Location** The magnetic field at a distance \( r \) from a long straight wire carrying current \( I \) is: \[ B = \frac{\mu_ I}{2\pi r} \] Given: - \( I = 70\,\text{A} \) - \( r = .20\,\text{m} \) - \( \mu_ = 4\pi \times 10^{-7}\,\text{T}\cdot\text{m/A} \) Plug in the values: \[ B = \frac{4\pi \times 10^{-7} \times 70}{2\pi \times .20} \] \[ B = \frac{4 \times 10^{-7} \times 70}{.40} \] \[ B = \frac{280 \times 10^{-7}}{.40} \] \[ B = 700 \times 10^{-7} \] \[ B = 7 \times 10^{-5}\,\text{T} \] **Explanation:** This step finds the strength of the magnetic field at the position of the moving bar using the Biot–Savart law for a straight wire. --- ### **Step 2: Calculate the Induced EMF and Current** The induced EMF (\( \mathcal{E} \)) in a moving conductor is: \[ \mathcal{E} = B \ell v \] where: - \( B = 7 \times 10^{-5}\,\text{T} \) (from Step 1) - \( \ell = .20\,\text{m} \) (distance between rails) - \( v = .90\,\text{m/s} \) \[ \mathcal{E} = (7 \times 10^{-5}) \times .20 \times .90 \] \[ \mathcal{E} = 7 \times 10^{-5} \times .18 \] \[ \mathcal{E} = 1.26 \times 10^{-5}\,\text{V} \] The resistor is \( R = 20\,\Omega \), so the current is: \[ I = \frac{\mathcal{E}}{R} = \frac{1.26 \times 10^{-5}}{20} \] \[ I = 6.3 \times 10^{-7}\,\text{A} \] **Direction:** By Lenz's Law, the induced current opposes the change causing it. The bar moves right, so the flux through the loop increases out of the page. The induced current will flow to produce a field into the page, which (using the right-hand rule) means **current flows from point a to b through the resistor**. --- ## **Final Answer** **The induced current in the resistor is \( 6.3 \times 10^{-7}\,\text{A} \), flowing from a to b.**