A pair of long, rigid metal rods, each of length 0.50 m, lie parallel to each other on a fricti

A pair of long, rigid metal rods, each of length 0.50 m,...

A pair of long, rigid metal rods, each of length $0.50 \mathrm{~m}$, lie parallel to each other on a frictionless table. Their ends are connected by identical, very lightweight conducting springs with unstretched length $l_{0}$ and force constant $k$ (Fig. $\mathbf{P 2 8 . 7 4}$ ). When a current $I$ runs through the circuit consisting of the rods and springs, the springs stretch. You measure the distance $x$ each spring stretches for certain values of $I$. When $I=8.05$ A, you measure that $x=0.40 \mathrm{~cm} .$ When $I=13.1 \mathrm{~A},$ you find $x=0.80 \mathrm{~cm} .$ In both cases the rods are much longer than the stretched springs, so it is accurate to use Eq. (28.11) for two infinitely long, parallel conductors. (a) From these two measurements, calculate $l_{0}$ and $k$. (b) If $I=12.0$ A, what distance $x$ will each spring stretch? (c) What current is required for each spring to stretch $1.00 \mathrm{~cm} ?$

Key Concepts

Magnetic Force Between Parallel Conductors

This concept describes the interaction between two long, current?carrying conductors. According to Ampère’s force law, parallel currents attract when flowing in the same direction and repel when in opposite directions, with the magnitude of the force per unit length given by an expression that includes the permeability of free space, the currents, and the separation between the conductors. This principle is central to understanding how electromagnetic forces act on current?carrying rods or wires.

Hooke's Law in Elasticity

Hooke’s law relates the force exerted by a spring to its displacement from the equilibrium (unstretched) position, expressed as F = kx, where k is the spring constant and x is the extension. This linear relationship is valid for small deformations and serves as a model for the elastic restoring force exerted by the springs in mechanical systems.

Force Equilibrium Analysis

In many physics problems, a system reaches equilibrium when forces acting in opposite directions balance each other out, resulting in zero net force. In the context of this problem, the magnetic force between the current?carrying rods is balanced by the elastic restoring force of the springs. This equilibrium condition allows for the derivation of relationships between the measurable quantities (such as spring stretch and current) and the physical constants of the system.

Experimental Determination of Physical Constants

This concept involves using experimental data to determine unknown parameters such as the spring constant and the unstretched length of the spring. By measuring the stretch of the spring under different currents, one can set up equations based on the balance between magnetic and elastic forces and solve for the unknowns. This approach illustrates the practical method of applying theoretical models to obtain quantitative information from experiments.

Transcript

00:01 Here we've got two parallel wires hooked to a circuit so that they both are hooked in series, and the current runs also through connected springs at both ends of the wires.

00:17 These are identical springs.

00:20 What we'll be using is some data, two data points on the current that flows through the wires versus the amount that the springs stretch.

00:34 And we see that they stretch because as the wires get occurred through them, they actually repel each other.

00:42 And so those two springs will stretch as the wires repel, and they will become of length l0 plus x.

00:54 What we'll be trying to do with that data is see if we can use the data to experimentally determine both the unstretched length of the springs and the spring constant of the springs, the identical springs.

01:16 There are some principles that we're going to use.

01:18 There's a number of them, but the first one is that the springs will stretch until they balance the magnetic force of repulsion from the current with the spring force.

01:38 So that is the main principle that we'll be using.

01:44 So we're after the two parameters in the spring, and we'll be using some known things about the magnetic force to evaluate those constants along with the data.

02:02 So what that means is we'll need to break apart each of those two forces.

02:08 And the force of the spring, i'll write down its magnitude, is the effective spring constant times l0 plus x.

02:21 This is the magnitude only.

02:23 The direction, of course, will be trying to pull back against the magnetic force.

02:30 We can kind of show that with some arrows.

02:42 There.

02:43 And the magnetic force will be equal and opposite on both wires, so we really need to just consider, say, the top wire.

02:54 And why i wrote it as effective spring constant is i do note that there are two springs.

03:00 They're working in parallel.

03:04 And so the spring stiffness effective is going to be bigger than just the single springs.

03:11 And since they're identical, it is twice.

03:16 The individual spring constant.

03:24 Okay, there are a couple things that i need to evaluate the magnetic force.

03:29 First of all, let's just evaluate the force on the top wire.

03:34 We'll call that wire one.

03:36 The force is going to come from the current in the wire times its length, which is half a meter, times the magnetic field that it's sitting in.

03:51 So magnetic field is coming from the bottom wire.

03:56 We'll call that wire 2.

04:00 And furthermore, that magnetic field can be evaluated using amper's law for a long straight wire.

04:12 We have nuo .i over 2 pi times the distance between them, which is l0 plus x.

04:24 So we can put that all together.

04:26 U -0 .i -squared length over 2 pi l -0 plus x.

04:41 Okay, so we have our two expressions now.

04:44 We have the magnetic force and we have the spring force.

04:50 And now let's equate them and see what we can do with our data to try to evaluate.

04:57 The stiffness and unstretched length.

05:03 Okay, so we're ready to put everything together.

05:12 So we have, let's go ahead and put it in red, mu nod i squared l over 2 pi l0 plus x equals k effective times l0 plus x.

05:36 And now i want to think about the data points that we're going to use.

05:42 What are the things that depend on that data? there's the current, and there is the x.

05:52 So what i'm going to do is i'm going to regroup this equation to put all the constants on one side of the equation and all the variables together on the other.

06:05 And the constants should be the same regardless of which data point you have.

06:15 So kind of rearranging, put the constants on the, i don't know, the right and the variables on the left.

06:41 I hope i don't mess up all these constants.

06:44 I tend to drop a few here and there.

06:50 But i do believe that looks like we have all of our constants together.

06:58 Okay, so what this says is that no matter what the data point is, so let's call these two data points maybe one and two.

07:16 I know i've got a lot of number ones and number twos, but here's data point one and data point two.

07:27 We know that they must agree because they concur in the constant on the right -hand side.

07:34 So what that means is, i for data point 1 squared, l -0 plus x1 quantity squared, equals i -2 quantity squared, l -0 plus x2 quantity squared.

07:58 Okay, and so notice there's only one unknown in that equation.

08:04 We've got two data points that we can put in.

08:07 If we wanted to, we could take the square root of both sides that might make things a little bit easier...

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