Video: Physics Past Exam • 2017/2018 • Pack 1 • Question 26

Physics Past Exam • 2017/2018 • Pack 1 • Question 26


Video Transcript

A rectangular coil of wire has 𝑁 turns and a surface area of 12.15 times 10 to the negative third square metres. The coil carries a current of three amps and is placed in a 0.4 tesla uniform magnetic field. The table below shows corresponding values of the magnetic torque 𝜏 on the coil and the sine of the angle 𝜃 between the normal to the coil and the direction of the magnetic flux lines. Draw a graph with 𝜏 values on the vertical axis and sine of 𝜃 values on the horizontal axis. From the graph, find the capital 𝑁, the number of turns in the coil.

This exercise then is all about a rectangular coil of wire. Let’s clear some space and draw a sketch of the coil. If we looked at our coil of wire side on, it might look like this: it has a cross-sectional area 𝐴 which is given to us and it carries a current 𝐼 in its loops also given. This coil as well sits in a uniform magnetic field we’ve called capital 𝐵.

If we were to rotate this coil 90 degrees so that we were looking at it from the side, we would see that it consists of some number of coils and that number is capital 𝑁 and that’s what we want to solve for. To get started solving for 𝑁, we’re going to consider the values in this table given to us.

To give some context for the values in this table, we know that our current carrying loops of wire are in a uniform magnetic field. That means this wire will experience a torque as it rotates. It’s that magnetic torque exerted on this coil of wire, which is given in this top row of values 𝜏 times 10 to the negative one newton metres.

And then in the bottom row, we have the sine of an angle 𝜃 where 𝜃 is the angle between the normal to our loop of wire and the magnetic field direction. As the angle 𝜃 between our loop and the external magnetic field changes, this table shows us how the torque on the current carrying wire changes as well.

Our next step will be to plot these values on a pair of axes with torque on a vertical axis and sine of 𝜃 on a horizontal axis. Here we have our axes drawn in, where on the vertical axis we show 𝜏 times 10 to the negative first newton metres ranging in values from zero to seven. And then, on the horizontal axis, we plot out the sine of the angle 𝜃 from zero all the way up to one in increments of one-tenth.

Our task now is to plot the values in this table on this graph that we’ve created. We’ll start out with the first point, where 0.2 is the sin of 𝜃 and 1.4 is 𝜏 times 10 to the negative first newton metres. We find 0.2 on our horizontal axis and 1.4 on our vertical axis and plot our point where the horizontal and vertical lines from these values intersect.

Then, we move on to our next point 0.4 and 2.8. Once again, we’ll look on our horizontal axis and find the value for sine 𝜃. And then, on our vertical axis, we’ll locate the value for 𝜏. And where these two points intersect in our graph, we’ll draw our data point. And we will continue on in the same process plotting our third, then our fourth, and finally our fifth and last data point.

Now that all our data is plotted here, we want to create a line of best fit that moves through the set of data when we create that line two things stand out. First, our line of best fit passes through the origin, the point zero, zero.

But much more unusual than that all five of these points are literally on the same line; they’re collinear with one another. This means that our line of best fit rather than passing near or close to some of the points actually lies directly on all of them.

At this point, we can recall that our problem statement directed us to solve for the slope of this line in order to help us solve for capital 𝑁, the number of turns in our coil. To find the slope of our line of best fit, that involves choosing two points on the line.

As one of our points, let’s pick the origin since that makes the arithmetic easier for our calculation. For our second point, we can take advantage of the unusual fact that all five of the points on our data table actually lie directly on our line. Therefore, they are points of our line of best fit.

So we can choose one of these five points to be our second point in the slope calculation. Let’s choose the point with coordinates 0.4, 2.8. So then to calculate our slope from our line of best fit, we’re using our second data point plotted as well as the origin.

When we enter those values into an equation for slope, being careful to account for the units of our torque values, we see that it’s equal to 0.28 minus zero newton metres all divided by 0.40[0.4] minus zero. This simplifies to 0.28 newton metres over 0.4. And our slope then is equal to 0.7 newton metres.

We’ve now solved for the slope. But how does that get us closer to solving for 𝑁, the number of turns in our coil? To see how, let’s recall the mathematical equation for the torque acting on a current carrying coil of wire in a uniform magnetic field.

That torque is equal to the number of turns in the coil times the current in the coil multiplied by its cross-sectional area times the strength of the uniform magnetic field it’s in all multiplied by the sine of 𝜃, where 𝜃 is the same as the angle 𝜃 in our example.

This equation is encouraging because it includes capital 𝑁, which is something we want to solve for. If we rearrange this equation to solve for capital 𝑁, we see it’s equal to 𝜏 divided by the sine of 𝜃 all multiplied by one over 𝐼𝐴𝐵.

When we write the equation this way, something may stand out: 𝜏 over the sine of 𝜃 is equal to the slope that we just calculated. After all, on the vertical axis, we had torque and on our horizontal axis we had the sine of 𝜃.

So this then is how our slope connects with the number of turns in our coil. It’s equal to 𝜏 over sine of 𝜃. And we can make this substitution: 𝑁 then is equal to the slope of our best fit line divided by the current in our coil multiplied by its area times the strength of the field that it’s in.

We’ve already solved for the slope of that line and we’re given 𝐴 and 𝐼 and 𝐵. So we’re to plug in and solve for 𝑁. When we enter this expression on our calculator, we find that 𝑁 is 48. That’s the number of loops or turns in our current carrying coil.

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