Two long, parallel wires hang by 4.00-cm -long cords from a common axis (Fig. P28.61). The wires

Two long, parallel wires hang by 4.00-cm -long cords from a...

Key Concepts

Magnetic Force Between Parallel Currents

This concept refers to the force that two current?carrying wires exert on each other as a result of the magnetic fields they produce. According to Ampère’s force law, the force per unit length between two parallel wires is proportional to the product of their currents and inversely proportional to the distance between them. This interaction is so fundamental that it serves in the definition of the ampere, the SI unit of current.

Magnetic Field of a Current-Carrying Conductor

A long, straight conductor carrying an electric current creates a magnetic field that encircles the wire. The strength of this field at a given distance from the wire is described by the Biot–Savart law, which shows that the field decreases with distance. Understanding this magnetic field is crucial for calculating forces between wires and for many applications in electromagnetism.

Equilibrium of Forces in a Suspended System

When a wire is suspended from a cord, it is subjected to multiple forces, chiefly gravitational and magnetic forces. In equilibrium, the vector sum of these forces is zero. By decomposing the tension in the supporting cord into vertical and horizontal components, one can set up equations balancing the gravitational force (vertical) and the magnetic force (horizontal). This technique of force resolution is a core principle in classical mechanics.

Force Decomposition and Tension Analysis

This concept involves breaking down a force (such as the tension in a cord) into its components along defined axes to analyze the equilibrium conditions. In problems involving suspended objects, the vertical component of tension typically balances the object's weight, while the horizontal component balances any lateral forces, such as the magnetic repulsion between current-carrying wires. This method provides a systematic way to solve for unknown quantities in a system in equilibrium.

Transcript

00:01 So a reminder about the magnetic field of a long straight wire that this may be found using amper's law.

00:10 And the magnetic field actually circulates in circles with the wire at the center, as shown in this little figure over to the right.

00:21 And the magnitude of the magnetic field drops off with the inverse of the distance from the wire.

00:29 So we're going to actually take a look at an example where we have two wires.

00:36 One is carrying a current in one direction, and the other is carrying a current in the opposite direction.

00:49 And we're going to examine the force between those two wires.

00:54 So the force is equal to i, and then the unit vector l is in the direction of the current itself, but l is the length of the wire and then crossed b.

01:10 So we'll call the wire on the left, wire number one, the wire on the right, number two, and they're separated by a distance.

01:20 We'll call that d.

01:23 And we're going to ask ourselves, what is the force between those two wires due to the magnetic field of one of the wires? so we're going to take a look at magnetic field of wire one is going to circulate around that wire.

01:37 And a counterclockwise sense, which means that at wire number two, the magnetic field is pointing up.

01:49 And the forestal number two, of course, is going to be equal and opposite to that on number one.

01:57 And it's equal to i -1, l -2, and then we have magnetic field b -1.

02:07 And if we look at i -cross -b, il cross b, the direction is outwards, which means that the force on the opposite wire is going to go in the opposite direction.

02:29 So we have a situation of repulsion.

02:36 And let's see, b1 is equal to i1, new not, over 2 pi d.

02:48 So the force on each wire is going to be proportional to the current.

02:53 In the other wire squared.

03:04 And that should have been i2.

03:13 Again, all my parameters straight.

03:17 I can look a little bit clearer, so i don't make that mistake again.

03:20 So, yeah, we have i2.

03:22 I1 is the product of the currents.

03:26 And then l2, u0 over 2 pi d, is the fourth between the two wires, and it must be equal and opposite on the, the other wire.

03:43 So this is a very common thing if you have an interaction between two objects at newton's third law, the characteristic that the property that's causing that force must show up as a product between the two objects.

04:05 So let's take a little further step.

04:11 The analogy to two charged objects hanging next to each other and repelling is two wires carrying current in opposite directions.

04:25 We're going to let them have the same current just for the simplification.

04:31 And we're going to make the wires very long so that the lengths don't really matter.

04:36 And we'll see how the lengths divide out.

04:38 But there's going to be a mass per unit length of 0 .0130 kilograms per meter for each wire.

04:49 And they're going to hang from a common thread.

04:51 By a little length of string l, and they're going to hang from the vertical at some known angle that you can measure theta, we'll call it.

05:09 And we're going to use that theta, the known theta, to determine the current.

05:23 So we'll start off with the current in terms of parameters given, and then put in some numerators given, quantities.

05:38 So we'll take a look at one of the two wires, maybe wire number two, since we have a picture up thereof.

05:48 So i2 is going to be a magnetic force to the right.

06:00 There will be mg downwards, and there will be a tension in the string going upwards.

06:17 And just like in the electrostatic situation, what we're going to have.

06:21 Have is that the tangent of theta is equal to tx over t y and t x is equal to the magnetic force and t y is equal to m g and we'll write that in terms of the mass per unit length times the length of the wire times g i'm not doing very good with my greek letter, but that's a greek letter z essentially.

07:08 Okay, so then we just simply use our f magnetic from above f magnetic...

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