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Question Video: Determining a Dimension of a Rectangular Loop of Wire in a Uniform Magnetic Field Physics

The diagram shows a rectangular conducting coil with 4 turns that is in a magnetic field with a strength of 325 mT. There is a current of 4.8 A in the coil. The sides of the loop parallel to line 𝑑₁ are parallel to the magnetic field. And the sides of the loop parallel to line 𝑑₂ are perpendicular to the magnetic field. The ratio of 𝑑₁ to 𝑑₂ is 1.5. The torque on the loop is 12.5 mNβ‹…m. Find the length of 𝑑₁ to the nearest millimeter.

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Video Transcript

The diagram shows a rectangular conducting coil with four turns that is in a magnetic field with a strength of 325 milliteslas. There is a current of 4.8 amperes in the coil. The sides of the loop parallel to line 𝑑 one are parallel to the magnetic field. And the sides of the loop parallel to line 𝑑 two are perpendicular to the magnetic field. The ratio of 𝑑 one to 𝑑 two is 1.5. The torque on the loop is 12.5 millinewton-meters. Find the length of 𝑑 one to the nearest millimeter.

In our diagram, we see our current-carrying conducting loop. This loop is a dimension 𝑑 one wide and a dimension 𝑑 two deep. It sits in between the poles of a permanent magnet. Therefore, the loop is exposed to a constant magnetic field pointing from the north to the south pole of the magnet. We’ll call this field 𝐡. Because this loop carries current and exists in a magnetic field, it experiences a torque. We’ll call that torque 𝜏. Along with all this, we’re told the current in the coil, we’ll call that value 𝐼. And we’re also told that the ratio of 𝑑 one to 𝑑 two, 𝑑 one divided by 𝑑 two, is 1.5. To complete our information, we’ll write that 𝐡 is 325 milliteslas and that the number of turns in our rectangular coil is capital 𝑁.

We can now clear some space to work and begin on our solution by recalling the general equation for the torque that acts on a current-carrying conductor in a uniform magnetic field. That torque is equal to the magnetic field magnitude multiplied by the current in the loop times the area of the loop multiplied by the number of turns in the rectangular coil times the sin of this angle πœƒ. If we were to look at a rectangular loop of wire from the side, as it’s oriented with the external magnetic field, then if we were to draw a vector that is perpendicular to the area of that rectangle, the angle between that vector and the external magnetic field is πœƒ. Notice that for our given rectangular coil, a vector normal or perpendicular to its surface meets the magnetic field lines at an angle of 90 degrees.

As we apply this general equation then, 𝜏 is equal to 𝐡 times 𝐼 times 𝐴 times 𝑁 times the sin of 90 degrees. And since the sin of 90 degrees equals one, we can write a simplified version of this equation. Let’s recall at this point that it’s the dimension of our rectangle 𝑑 one that we want to solve for. That’s the larger of the two dimensions of our rectangle here. That dimension isn’t in this equation directly. But because our coil is in the shape of a rectangle, 𝐴 is equal to 𝑑 one times 𝑑 two. So torque 𝜏 equals 𝐡 times 𝐼 times 𝑑 one times 𝑑 two times 𝑁.

And now we’ll make use of the fact that the ratio of 𝑑 one to 𝑑 two is 1.5. If we write this out a bit differently, we see that 𝑑 one divided by 𝑑 two equals 1.5 divided by one. If we multiply both sides of this equation by 𝑑 two so that that factor cancels on the left and if we then multiply both sides of the equation by one divided by 1.5, then over on the right-hand side, 1.5 divided by one cancels with one divided by 1.5. We’re left with 𝑑 two on the right. And on the left, we get 𝑑 one divided by 1.5. It’s useful to know this because now we can replace this 𝑑 two in our equation with 𝑑 one divided by 1.5.

Looking at our overall equation now, we see that we’re really making progress. We’re given the value of the torque 𝜏. We know the magnetic field strength 𝐡, the current 𝐼, as well as the number of turns in our coil 𝑁. Another way to write the right-hand side of this expression is as 𝐡 times 𝐼 times 𝑁 over 1.5 all multiplied by 𝑑 one squared. What we can do as a next step is take this whole equation and multiply both sides by 1.5 divided by 𝐡 times 𝐼 times 𝑁. Over on the right-hand side, this completely cancels out 𝐡, 𝐼, 𝑁, and 1.5.

As a final rearrangement, we can take the square roots of both sides of this expression. Taking the square root of 𝑑 one squared gives us 𝑑 one by itself. Flipping this equation around, we now have an expression for 𝑑 one. And remember, we know the values of the torque 𝜏, the magnetic field strength 𝐡, the current 𝐼, and the number of turns in our coil 𝑁. With these values plugged in, we just want to make a couple more changes before we calculate 𝑑 one.

Notice that in the denominator, the units of our magnetic field are milliteslas. We’d like to convert those units to teslas. And we can do so by recalling that 1000 milliteslas equals one tesla. To make the unit switch then, we’ll move the decimal place in this number one, two, three spots to the left. 325 milliteslas equals 0.325 teslas. Then, in the numerator, with our millinewton-meters, we’ll do something similar. 1000 millinewton-meters equals one newton-meter. So our torque of 12.5 millinewton-meters equals 0.0125 newton-meters.

We’re finally ready to enter this expression on our calculator and compute 𝑑 one. The result we get looks like this. And it’s important to note that the units of this result are meters. However, we want our final answer to be rounded to the nearest millimeter. To make that switch, we’ll move the decimal place in our answer one, two, three spots to the right. So we now have 54.81 and so on millimeters. And then we’ll round this result to the nearest whole number. Since the first digit after the decimal is greater than or equal to five, our answer comes to 55 millimeters. This is the length of the dimension 𝑑 one rounded to the nearest millimeter.

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