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

A pair of long, rigid metal rods, each of length 0.50 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 $k$ ($\textbf{Fig. P28.78}$). 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 cm. When $I =$ 13.1 A, you find $x =$ 0.80 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 cm?

Transcript

00:01 Okay, this is question 28 .78.

00:03 We have a pair of long, rigid metal rods here, each of length 0 .5 meters, which are parallel to each other on a frictionless table.

00:12 And they're connected by these two springs.

00:16 Now, the springs have unstretched length l0, which i've actually just relabeled here, j, because it's hard to see.

00:23 The l0 looks a little bit like a 10 here.

00:24 So i'm going to just say unstretched length l0 or 10.

00:28 And force constant k.

00:31 And we're told, we have these two data points.

00:34 We're told that when a current i runs through the circuit, the springs stretch by a distance of 0 .04 meters, or 0 .4 centimeters, which i've converted to meters here, so 0 .004 meters.

00:46 And when a current of 13 .1 amps runs through this, we get a stretch here of 0 .008 meters, 0 .8 centimeters.

00:55 So if we express this total distance d here between the two rods, then we can express this total d by j or l0, that's this unstretched length, plus x, which is the amount that it's going to stretch.

01:11 So this is what we're going to use.

01:13 So in order to solve this, what we're going to need to do is think about the force between these two rods due to the magnetic field from the current.

01:24 That force is going to be pushing the two rods away since these currents are running in opposite directions.

01:31 And then the force of the spring is going to be pulling the two rods together.

01:36 And so what we want, since we know that this whole system isn't moving, we want to set the force from the magnetic field equal to the force on the spring, coming from the springs.

01:46 So that will mean that our net force is zero, which is the state that our system is in here.

01:54 So let's first take a look at the force between two wires.

01:58 This is the general expression that we have here for the force per unit length of two wires with current i1 and i2.

02:07 Let's just take a look here.

02:08 We want just the force.

02:09 We don't want force per unit length.

02:11 So let's start by just multiplying by the length on both sides.

02:15 That's just going to bring this over to the numerator here.

02:19 Now i1 and i2 are actually both just equal to i because the same current.

02:24 Is running through these two wires.

02:26 So i can turn this i -1 times i -2 just into i -squared up here.

02:32 And now r is the only other thing we don't know.

02:35 And r is the distance between the two wires.

02:39 So the distance between these two rods here is going to be equal to d.

02:43 So we're going to just set d there.

02:46 And now the force from a spring, the force in general is going to be k times d.

02:52 This is the force of a spring with constant k stretching over a distance d.

02:58 But we have two of these springs acting on each rod.

03:02 So i'm going to multiply that by two.

03:03 So we have 2kd is going to be the force pulling these two rods together.

03:10 And this is the magnetic force pushing them apart.

03:13 So now what we want to do is set these equal.

03:16 So i'm going to come over here and set this.

03:22 Equal to 2kd.

03:26 And now we just want to rearrange this and simplify this a bit.

03:31 So let's keep our mu knot i squared over here and multiply by 2 pi d on both sides.

03:39 That's going to give us a 4 pi kd squared.

03:45 And let's solve this for d.

03:47 So if we take d squared, switching this the order here, i'm going to we're going to have this in the numerator divided by our 4kd, our 4 pi k, i'm sorry, in the denominator.

04:05 And so now if we want to solve for d, we're going to need to just take the square root of both sides.

04:11 When we take the square root of this right side, we see we have an i squared, so we can pull one factor of i out.

04:19 And we'll be left with this, just one second.

04:24 I'm going to draw the square root right here.

04:30 So this whole thing, taking the square root.

04:35 So now that we have this, all right, remember that our d is equal to j plus x, and we want to solve for j.

04:46 J is our l not here.

04:48 So i'm gonna put in j plus x in for d and that's equal to i.

04:54 And at this point i'm just gonna use the power of one half instead of writing out this whole square root sign.

05:03 Taking the square root is equal to raising to a power of one half.

05:08 So i'm just going to write this just like this.

05:13 So now let's take a look at what we have.

05:15 For each value that we're told here, we're given i and we're given x.

05:22 So let me just highlight this here.

05:26 So we're given x and i for some given each of these points.

05:32 And what we want to solve for is l -not or j and k.

05:37 These are the two values that part a is asking us to find.

05:41 So what we're going to do is we're going to plug each of these values in.

05:44 And then that's going to give us two equations, and we'll have two unknowns, our j and our k.

05:50 And we're going to have to solve that system of equations.

05:53 So i'm just going to do that down here.

05:57 I'm just going to plug in the two values that we know so that we can have our equations to solve...

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