The long, straight wire shown in Fig. 29.51 a carries constant current I . A metal bar with leng

The long, straight wire shown in Fig. 29.51 a carries...

Key Concepts

Ohm’s Law in the context of Induced Currents

Ohm’s Law relates the current flowing through a conductor to the voltage (or emf) across it and the resistance of the conductor. In systems where an induced emf drives a current through a resistive loop, this law allows one to calculate the magnitude of the induced current by dividing the emf by the resistance, thus linking electromagnetic induction with circuit behavior.

Lenz’s Law

Lenz's Law states that the direction of the induced current (or emf) in a circuit will be such that its own magnetic field opposes the change in the magnetic flux that produced it. This law provides a qualitative way to determine the direction of induced currents and ensures the conservation of energy in electromagnetic processes.

Motional Electromotive Force (EMF)

Motional emf is generated when a conductor moves through a magnetic field, causing free charges within the conductor to experience a Lorentz force. This force results from the motion of the conductor in the presence of the magnetic field, leading to a separation of charges and hence a potential difference along the conductor. The emf is typically calculated as the product of the velocity, magnetic field strength, and effective length of the conductor perpendicular to the field.

Magnetic Field of a Current-Carrying Conductor

A steady current produces a magnetic field in the space surrounding the conductor, with the field lines forming concentric circles around the wire. The strength of this magnetic field depends on the magnitude of the current and the distance from the wire, typically described by the Biot–Savart law or Ampere’s law, and is essential for determining the magnetic flux in problems involving induction.

Faraday’s Law of Induction

This principle states that a change in magnetic flux through a closed loop induces an electromotive force (emf) in the circuit. It forms the foundation for understanding how electrical energy can be generated from mechanical motion, and it mathematically relates the emf to the rate of change of the magnetic flux through the surface bounded by a loop.

Transcript

00:01 In this question, we are going to determine the induced emf and induced current respectively for a bar moving near a current -carrying wire and for a loop moving near a current -carrying wire.

00:19 To find the emf induced in the bar, we need to begin with the equation for motional emf, which is the line integral of the cross product of the velocity and magnetic field times dl, where dl is the length of our conductor that's going to be parallel to the magnetic force.

00:59 So what's really nice here is that, excuse me, so v and b are already perpendicular, so we can simply take v times b.

01:14 And then the thing that's going to be a little bit tricky here is that we have a spatial variation of the magnetic field because the magnetic field due to a current -carrying wire is mu -naught i over 2 pi r, where r is the distance away from the wire.

01:34 So clearly b will be weaker at point b on the bar and it will be stronger at point a.

01:51 So i mean, this is all going to work out, don't worry.

01:54 So what ends up happening here is that dl ends up becoming a differential for how b varies in space.

02:32 And so we can now say that our emf will be equal to the line integral.

02:39 And now here's the thing, i'm going to start from point b because our magnetic field due to this current -carrying wire, according to the right -hand rule for the field due to a current -carrying wire, is going to be into the page and it's going to be very strong at position a and it's going to be less strong at in the middle of the bar and then it's going to be weaker over here at position b.

03:13 So the right -hand rule for a charged particle moving in a magnetic field tells me that regardless of the strength of my field, the magnetic force on these charged particles is going to be upward.

03:34 So i'm going to start my integration from point b and integrate up to point a so that f and dl are in the same direction.

03:46 So we will be integrating from d plus l, where d is the distance from the wire and l is the length of the bar, up to d of v, the velocity given, times mu naught i over 2 pi.

04:06 And i'm going to go ahead and just use little l as my integration variable times dl.

04:14 And notice that everything except for my 1 over l comes out.

04:25 So v mu naught i over 2 pi times the integral from d plus l to d of 1 over l dl is my result of this...

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