### Electromagnetic Induction in a Moving Conductor**Setup Description:**Two identical conducting bars, each with a length of 56.6 cm, can move across two parallel conducting wires. The bars can either move to the left or right with velocities \( v_1 \) and \( v_2 \). The top wire is equipped with a resistor (R) having a resistance of 3.2 Ω, while the bottom wire includes an ammeter (A) to measure the current. The ammeter's resistance is negligible and reads positive when the current flows to the right.A uniform magnetic field (\( \mathbf{B} \)), with strength \( B \), is applied perpendicular to the plane of the wires and bars, either pointing into or out of the page (indicated by symbols ⊗ or ⨀).**Diagram Explanation:**- **Bars and Wires:** Two parallel horizontal conducting wires with a vertical space where the bars slide. Located between the bars is the magnetic field.- **Components:** - **R:** Resistor attached to the top wire. - **A:** Ammeter on the bottom wire indicating current flow. - **\( B \) Field:** Shown by the direction symbols (in or out of the page).**Scenario 1:**- **Conditions:** The right bar remains stationary, while the left bar moves rightward with a constant speed \( v_1 = 4.1 \, \text{m/s} \).- **Magnetic Field:** Directed into the page with a strength of \( 6.2 \, \text{T} \).**Problem 1:** Calculate the Electromotive Force (EMF) induced in the left bar. Note that a positive EMF indicates the top of the bar is at a higher potential than the bottom.**Problem 2:** Determine the current measured by the ammeter for scenario 1.**Submission Details:**- Each problem allows up to 10 submissions.- Current submission count: 0 for both questions.

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Answered: Two identical conducting bars of length…

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