A bar of length L=0.36 m is free to slide without friction on horizontal rails as shown in Fig.

A bar of length L=0.36 m is free to slide without friction...

Key Concepts

Terminal Speed in Electromechanical Systems

Terminal speed is reached when the driving force (due to an applied voltage or other motive factors) is exactly balanced by the opposing induced forces (such as those from a motional EMF). At this point, the net force on the moving object is zero, so acceleration ceases and the object continues to move at a constant speed. This concept illustrates dynamic equilibrium where ongoing processes balance out, leading to steady-state conditions in the system.

Electromagnetic Induction

This concept refers to the generation of an electromotive force (EMF) in a conductor by a changing magnetic flux. When a conductor moves through a magnetic field, or when the magnetic field around a conductor changes, an induced EMF arises according to Faraday's law. Lenz's law further determines the direction of the induced current such that it opposes the change in flux, which is fundamental in understanding how motion and magnetic fields interact in electrical circuits.

Motional EMF

Motional EMF is a specific case of electromagnetic induction where the EMF is generated due to the motion of a conductor within a magnetic field. The induced voltage is proportional to the magnetic field strength, the velocity of the conductor, and the length of the conductor segment that is perpendicular to the direction of the magnetic field. This concept is key to understanding how mechanical motion can produce electrical effects in a circuit.

Lorentz Force on a Current-Carrying Conductor

A current-carrying conductor in a magnetic field experiences a force due to the Lorentz force law. This force is given by the cross product of the current element and the magnetic field, and it acts perpendicular to both the direction of the current and the magnetic field. This concept is important in electromechanical systems as it explains the interaction between electrical currents and magnetic fields that leads to mechanical motion.

Circuit Analysis with Induced EMF

In circuits where there is both an applied voltage from a battery and an induced EMF due to motion in a magnetic field, Kirchhoff's loop rule is used to analyze the net EMF and current in the circuit. The induced EMF, which opposes the applied voltage because of Lenz’s law, modifies the net current flowing through the circuit elements. This analysis is essential for determining current, voltage drops, and forces in systems that combine electrical circuits with mechanical motion.

Newton's Second Law in Electromechanical Systems

Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma). In electromechanical systems, the magnetic forces resulting from induced or applied currents are analyzed using Newton's second law to determine the resulting acceleration and motion. This law provides the connection between electrical forces and mechanical motion, enabling prediction of the dynamic behavior of the system.

Transcript

00:01 Hi there, so for this problem, and we have the situation that is shown here in this figure.

00:08 Now the length of this bar is equal to 0 .36 meters, we're also given the magnetic field that is equal to 2 .4 tesla.

00:21 And the nph that is applied to this is 12 balls.

00:26 And the bar has a mass that we're going to label as lowercase m that is 0 .4 .0 .000.

00:33 And the resistance for this also given that is 5 oms now for part a of this problem we are asked to sketch the bar's speed as a function of time okay so um then okay so we know that the net current in the loop is just the nph minus the period between the magnetic field inductance the speed and this, well, sorry, the length and the speed, and this divided by the resistance, okay? now the acceleration of the bar, we know that the acceleration is the force divided by the mass.

01:34 So we can substitute in here the magnetic force, which is the current, the length for this, the magnetic fill, this times the sign of the angle, which is the sign of 90 degrees, and this divided by the mass.

01:49 Now in here, we can substitute the expression that we know for the current.

01:55 So that will be that this is the l minus the magnetic field, the length, the speed.

02:09 This times the length, the magnetic feel, and this divided by the mass times the resistance.

02:20 So once we have thought, we need to fly the speed.

02:25 So we know that the acceleration is just the rate of change of the speed, with respect to time.

02:31 So we just need to simply integrate this.

02:36 So we will have that this is the integral from zero to the speed b of the differential in b.

02:42 This after separation of variables, of course.

02:47 And this is equal to the integral from zero to the times t of the length, the magnetic field, divided by the mass, times the resistance.

02:56 This integrated over the time.

02:58 So the solution for this is that the speed is equal, to the nph divided by the magnetic field, the length l, this, times one minus the exponential of minus the magnetic field to the square times the length to the square times the time.

03:18 This divided by the mass times the resistance.

03:21 So that's a solution for the speed.

03:27 We can substitute the values that we are given for this.

03:30 So after we do that, we'll do that.

03:32 We obtain 14 and this times 1 minus the exponential of minus the time divided by 6.

03:41 Okay, so we can do this.

03:44 Let me just do this graph in here.

03:48 Well, of course, you can use a graphic calculator and going to just draw the idea for this.

03:57 Okay, so you should obtain a graph in the following form that is at zero and increases at something like this.

04:10 Okay, so in here we should have the terminal velocity.

04:16 So the vertical axis is for this speed and the horizontal axis is for the time.

04:22 So that's a solution for part a of this problem.

04:26 Now for part b, the question is about just after the switch is closed, what is the acceleration of the bar? now, we know that just a question is about, after the switch is closed, the speed is zero...

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