As a summer intern at a research lab, you are given a long...

As a summer intern at a research lab, you are given a long solenoid that has two separate windings that are wound close together, in the same direction, on the same hollow cylindrical form. You must determine the number of turns in each winding. The solenoid has length $L =$ 40.0 cm and diameter 2.80 cm. You let a 2.00-mA current flow in winding 1 and vary the current I in winding 2; both currents flow in the same direction. Then you measure the magnetic-field magnitude $B$ at the center of the solenoid as a function of $I$. You plot your results as $BL/\mu_0$ versus $I$. The graph in $\textbf{Fig. P28.76}$ shows the best-fit straight line to your data. (a) Explain why the data plotted in this way should fall close to a straight line. (b) Use Fig. P28.76 to calculate $N_1$ and $N_2$, the number of turns in windings 1 and 2. (c) If the current in winding 1 remains 2.00 mA in its original direction and winding 2 has $I =$ 5.00 mA in the opposite direction, what is $B$ at the center of the solenoid?

Transcript

00:01 Okay, this is question 28 .76, where we have this graph, figure 2876, that we're going to need to reference.

00:09 And we have a solenoid that has two separate wires that we're winding around.

00:15 And the wires are wound in the same direction.

00:18 We have the same cylinder that they're being wound around.

00:23 The length of the solenoid is 40 centimeters, a 0 .4 meters.

00:29 And we're told the diameter as well.

00:32 And we're told that one of the currents in one of the wires is going to be fixed at 2 millimet amps, which is 0 .002 amps.

00:40 And what we're trying to figure out is what are the number of windings in the first, in wire one and in wire two.

00:50 So this is a magnetic field if we had just a regular solenoid with only one wire, n is the number of windings, and i is a current in that one wire.

01:02 When we have two, here i'm just going to, we can just take the sum, since the magnetic field they're just going to add, we'll have this expression for the first one, and then we'll have the number of bindings on the second wire plus the current in the second wire.

01:21 This is going to be the total net magnetic field from two wires that are wound around creating a solenoid.

01:27 So we can rewrite this just a little bit.

01:33 We can pull out this factor of nuna over l in both of these terms.

01:38 Since those are both constants, this length isn't going to change.

01:42 And then we have n1 times i1 plus n2 times i2.

01:49 And so what we can do here, this is the axis, the y -axis in this figure that we're given, gives us a hint that we're going to want to multiply by l over mu not, which is going to bring this to the other side.

02:03 So we'll get b times l divided by mu not.

02:07 And that's great because that's going to give us our y component on the axis here.

02:12 And we have, remember, we're told what i -1 is.

02:18 So i'm just going to plug that in explicitly here.

02:21 I -1 is 2 millimet, so 0 .002.

02:26 Amps.

02:27 So i'm just going to plug that in explicitly so that we can see this is an unknown variable because we don't know b.

02:35 N1 we don't know and two we don't know and i2 we also don't know.

02:40 So now if we take a look at this graph that is plotted, basically what's going on is we're varying i2.

02:47 So we're going to change the current in i2 and see what happens to this over here.

02:53 Now, if we take a look at the form of this here, this is the equation for a regular line.

03:02 If we have the form y equals m times x plus b, because n1 and n2 really are constants here.

03:10 So here n2 is really going to be the slope.

03:12 That's going to be the equivalent of this m.

03:14 X is the variable that we're changing.

03:17 So that's i2.

03:18 And then we have this other constant here, 0 .002 times n1.

03:22 Again, n1 is just a constant we need to determine.

03:25 And that's our b.

03:26 And y, again, this is just the resulting variable that we're going to get.

03:30 So this is in the form of a straight line.

03:32 So that's why we expect to see this straight line when we plot this data.

03:39 So now we need to actually calculate the current in or the number of windings, one and two.

03:47 So we're going to want to use our graph here for this.

03:50 And so what a good thing to do is to look for points.

03:53 That look really easy to use.

03:56 So points where they lie right on an intersection, where we can really estimate a good value for them.

04:03 So the points that i'm going to use here are the first one where we have i2 is equal to 2 millimetps.

04:11 So that's going to be 0 .002 amps.

04:15 And when we have that, this factor bl over mu not is equal to, if we look at the value on the y -axis, that's equal to 8.

04:29 And the other point that looks nice to use is going to be when i -2 is equal to 5 millimet amps, then our bl over mu -not is equal to 16, because those both fly close to a little corner or intersection of the grid.

04:49 So it looks like it's easy to estimate.

04:51 So i'm going to pick those two points to use.

04:54 Theory, it doesn't really matter which points, but the ones that are in the middle of the lines are a little bit harder to estimate the exact value, so you may not get a great number when you use those.

05:05 So now what we want to do, we know that we're going to need two points because we have two unknown values here.

05:10 We don't know n1 and we don't know n2...

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