In the figure below, a metal bar sitting on 2 parallel conducting rails connected to each other by a resistor is pulled to the right at a constant speed. The resistance R=9 ohms, the distance between the rails is l=1.2m and a uniform 2.10T magnetic field is directed into the page. At what speed (m/s) should the bar be moved to produce a current of .500 A in the resistor? The diagram illustrates a visual representation involving a spring and an applied force. Here's a detailed explanation:1. **Components**: - **Spring (R)**: Depicted on the left side of the diagram, a spring is represented with a zigzag line. - **Distance (\(\ell\))**: An arrow labeled \(\ell\) runs vertically indicating a distance or displacement within the system. - **Block**: Positioned in front of the spring, there is a blue-shaded block. - **Applied Force (\(F_{\text{app}}\))**: An arrow pointing to the right from the block indicates the direction of the applied force, labeled \(F_{\text{app}}\).2. **Surroundings**: - The entire setup is placed between two horizontal surfaces, colored orange, indicating some form of boundary or constraint.3. **Function**: - The diagram appears to represent a mechanical system where a force is applied to move the block, compressing or extending a spring.This representation is useful for understanding fundamental concepts of mechanics such as Hooke's Law, forces, and energy transformation in physics.

Givens: Magnetic field, B=2.10 T Distance between the rails, l=1.2 m Resistance, R=9 Ω Current, i=0.500 AIf the speed of the rod is 'v' then the formula for induced emf in the rod is, e=Blv Hence, the current is: i=eR=BlvR Substituting the given values we get: i=BlvRv=iRBlv=0.500×92.10×1.2v=1.786 m/s Hence, the speed of the rod should be 1.786 m/s.