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 𝐵. 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 𝐵 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 𝜃. 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 𝜃. 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 𝐵 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 𝐵 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
𝐵, 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 𝐵 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 𝐵 times 𝐼 times
𝑁. Over on the right-hand side, this
completely cancels out 𝐵, 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 𝐵, 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.
