Two parallel rails with negligible resistance are 10.0 cm...

Two parallel rails with negligible resistance are $10.0 \mathrm{~cm}$ apart and are connected by a $5.00-\Omega$ resistor. The circuit also contains two metal rods having resistances of $10.0 \Omega$ and $15.0 \Omega$ sliding along the rails (Fig. P31.31). The rods are pulled away from the resistor at constant speeds $4.00 \mathrm{~m} / \mathrm{s}$ and $2.00 \mathrm{~m} / \mathrm{s}$, respectively. A uniform magnetic field of magnitude $0.0100 \mathrm{~T}$ is applied perpendicular to the plane of the rails. Determine the current in the $5.00-\Omega$ resistor.

Key Concepts

Motional Electromotive Force (EMF)

Motional EMF is the voltage induced when a conductor moves through a magnetic field. The magnitude of the induced EMF is given by the product of the magnetic field strength, the length of the conductor in the field, and its speed perpendicular to the field. This concept is crucial for understanding how moving conductors, like the metal rods in the problem, generate voltage and drive current in a circuit even in the absence of traditional batteries.

Faraday's Law of Electromagnetic Induction

Faraday’s Law states that a change in magnetic flux through a circuit induces an electromotive force (EMF) in that circuit. In cases where conductors move in a constant magnetic field, the change in their position alters the enclosed magnetic flux over time, leading to an induced voltage. This principle is central to analyzing circuits with moving parts, as it describes how mechanical motion can be converted into electrical energy.

Ohm's Law and Circuit Analysis

Ohm’s Law, which relates voltage, current, and resistance (I = V/R), is fundamental in determining the current that flows through various circuit elements when an EMF is present. In problems involving multiple resistances—including induced EMFs from moving conductors—circuit analysis techniques allow us to combine these resistances appropriately (whether in series or parallel) to calculate the resulting current in a specific component of the circuit, such as the resistor highlighted in the problem.

Transcript

00:01 There are two rails in the given problem at a distance of 10 cm and two rods are sliding over these rails.

00:20 These are the two roars and a resistance is also connected between the rails.

00:30 The rods are moving away from this resistance with the velocities v1 and v2 respectively and there is a magnetic field in the region which is directed into the plane of paper so both of these rods will intercept the magnetic field lines perpendicularly so emotionally will be induced across the ends of the rods so given is length of the rods which is equal to the gap between the rails means 10 .0 centimeter or we can say 0 .1 meter the velocity v1 for the first row is given as 4 .0 meter per second the velocity for the second rod is given as 2 .0 meter per second and the magnetic field into the plane of paper here this is given as 0 .01 tesla.

01:44 So first of all, we will find the motionally enf induced across the first drawer, which will be given as e1, is going to be b into v1 into l, 0 .01 into 4 into 0 .1 volt.

02:05 Or i can say 4 into 10 dash to bar minus 3 volt similarly for the second rod p into v 2 into l 0 .01 into 2 into 0 .1 volt or i can say 2 into 10 dash bar minus 3 volt the resistances of the rods are given as r1 and r2 and this is r3 these is r3 these values are as given here.

02:44 There are 1 is 10 om, r2 is 15 om and r3 is given as 5 ome.

03:00 Now as for the directions of these emfs are concerned using fleming's right hand rule the direction of this emf will be downward it will be sending the current like this in the direction of this emf here it is upward like this it will be sending the current upward now we rearrange the circuit as given we make the loops we replace the rods with the cells with the emf and their resistances will behave like the internal resistance of the cell so we make the circuit like this here this is r3 which is 5 -oam this is r1 which is 10 om and here this is r3 r2 which is 15 om here this emf induced is 4 into 10 dashed per minus 3 volt and this emf induced 2 into 10 dashed by minus 3 volt then we name the circuit as a b or yeah a b c b b f f f f f we name it now we show the current distributions in the loops suppose a current i1 comes out of it and a current i2 comes out of the second here so here this current then it will reach here at this point c according to critchov's junction rule it will be divided into two parts here this will be i2 as the current will have to remain same enough branch and this branch is already having a current i2 so here this is i2 so the remaining current i 1 minus i 2 will go there like this on reaching at f i 2 will join this i 1 minus i 2 and again it will become i 1 and will enter into this so now we will apply kirchhoff's voltage law first of all in closed loop a b c f a using kirchhoff's voltage law in closed loop a b, cfa, a, b, cfa, a, b, cfa...

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