Video Transcript
The diagram shows a rectangular
conducting coil with five turns that is in a magnetic field with a strength of 650
milliteslas. 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 length of 𝑑 one equals 0.055
meters, and the length of 𝑑 two equals 0.035 meters. The torque on the loop is 1.2
millinewton-meters. What is the current in the
loop? Answer to the nearest
milliampere.
As we get started with this
example, let’s write down some of the information we’re given. We’re told that a rectangular
conducting coil here has five turns. We’ll call that value capital
𝑁. We also know that the magnetic
field between the poles of this permanent magnet in which the coil exists, we’ll
call it 𝐵, is 650 milliteslas. The dimensions of our rectangular
coil are 𝑑 one and 𝑑 two. We’ll record those values at the
bottom of our page, and lastly, we’re told about the torque that acts on our
current-carrying loop. We’ll call that torque 𝜏, and it’s
given as 1.2 millinewton-meters.
Whenever we see this, we’ll have to
be careful to remember that this first m stands for milli-, the prefix, and the
second m stands for meters, the unit. Anyway, knowing all this, we want
to solve for the current in the loop. To get started, let’s clear some
space on screen. And let’s recall that, in general,
the torque that acts on a current-carrying loop of wire in a magnetic field equals
the magnetic field strength 𝐵 times the current in the wire 𝐼 multiplied by the
cross-sectional area of the loop times the number of turns in the loop all
multiplied by the sin of the angle 𝜃.
To understand 𝜃, say that we have
our current-carrying loop in our uniform magnetic field 𝐵. If we draw a vector that is
perpendicular to the plane of this loop, then the angle between that vector and the
external magnetic field 𝐵 is 𝜃. For the current-carrying loop we
have here, a vector that is normal or perpendicular to its surface points directly
upward. The angle between this vector and
the magnetic field is 90 degrees. Therefore, when we apply this
relationship, we’ll find that the sin of 𝜃, the sin of 90 degrees, is one. Knowing this, we can write this
simplified equation for our expression for torque.
We recall though that it’s not the
torque we want to solve for but rather the current in our conductor. If we divide both sides of this
equation by 𝐵 times 𝐴 times 𝑁, then those factors all cancel out on the
right. We then have an expression where
𝐼, the current, is the subject. Considering the variables on the
right-hand side of this expression, we know the torque 𝜏. That’s given to us in the problem
statement. We also know the magnetic field 𝐵
and the number of turns in our coil 𝑁. Concerning the area of our loop 𝐴,
we’re not directly told this value, but we do know the perpendicular dimensions of
the loop, 𝑑 one and 𝑑 two. The product of these two dimensions
equals the area 𝐴.
If we substitute in for 𝜏, 𝐵, 𝑑
one, 𝑑 two, and 𝑁, we’re just about ready to calculate and solve for the current
𝐼. Before we do though, we’d like to
change the units of our magnetic field from milliteslas to teslas. 1000 milliteslas equals one
tesla. So to convert between these two
units, we can shift the decimal place one, two, three spots to the left. 650 milliteslas is 0.650
teslas. Now, we could do something similar
in our numerator converting millinewton-meters to newton-meters. But in our question statement, we
were asked to solve for 𝐼 and round it to the nearest milliampere.
That means if we keep this one
milli- prefix on the right-hand side of our expression, then when we calculate this
whole fraction, we’ll get an answer in milliamperes. That brings us closer to our final
answer, so we’ll take that approach. This fraction is equal to 191.80
and so on milliamperes. And when we round this to the
nearest milliampere, eight, being greater than or equal to five, leads us to round
up so that our final answer is 192 milliamperes. This is the magnitude of the
current that exists in our conductor.
