### Video Transcript

The diagram shows a rectangular
loop of current-carrying wire between the poles of a magnet. The sides of the loop parallel to
line ๐ one are parallel to the magnetic field, and the sides of the loop parallel
to line ๐ two are perpendicular to the magnetic field. The current in the loop is 350
milliamperes, and the magnetic field strength is 0.12 tesla. The length of ๐ one equals 0.025
meters, and the length of ๐ two equals 0.015 meters. Find the torque acting on the loop
to the nearest micronewton-meter.

Looking at our diagram, here is our
current-carrying loop in a uniform magnetic field between the poles of this
magnet. Because thereโs current in all four
sides of our loop, there is potentially a magnetic force on each of these four
sides. In our problem statement though, we
were told that the two sides of the rectangle parallel to this distance ๐ one, so
that would be this side here and this side here, are both exactly parallel to the
external magnetic field. Therefore, the current in these two
sides of the rectangle is either parallel or antiparallel with the external
field. And that means these sides of the
rectangle experience no magnetic force. Itโs only the two sides
perpendicular to that field, this side and this side of the rectangle, that
experience a nonzero magnetic force.

We can determine the direction of
that force on each of these side lengths according to whatโs called a right-hand
rule. Given a wire carrying current weโll
call ๐ผ, in a uniform magnetic field weโll call ๐ต, then using our right hand, if we
point the index finger of that hand in the direction of ๐ผ and then the middle
finger of that hand in the direction of ๐ต, that will make the thumb of our right
hand point in the direction of the resultant magnetic force. That is, the force on this segment
of current-carrying wire will be out of the screen towards us. This right-hand rule will let us
figure out the direction of the force acting on either side of our rectangle thatโs
perpendicular to the external magnetic field.

On the side of the rectangle thatโs
to the left as we see it, this side here, pointing the index finger of a right hand
in the direction of current here and the middle finger of that hand in the direction
of the external magnetic field, left to right, the right-hand rule tells us that the
resulting magnetic force direction is down. Considering next the opposite side
of our rectangle, in this side, current points in this direction. Like before, the external magnetic
field points left to right so that now by our right-hand rule the force on this
segment of the rectangle is upward.

The combined effect of these two
forces is to create a torque. There will be a tendency of our
coil of wire to rotate about an axis through the loopโs center. This is the torque our question is
asking about. Thereโs a mathematical relationship
for the torque on a current-carrying wire in a magnetic field that we can
recall. That torque ๐ is equal to the
magnetic field strength ๐ต times the current in the wire ๐ผ multiplied by the
cross-sectional area of the wire ๐ด times the number of turns in the wire multiplied
by the sine of an angle weโve called ๐.

Before weโre able to evaluate this
expression for our scenario though, letโs recall the information given to us in our
problem statement. Our current-carrying loop, we
remember, has dimensions ๐ one multiplied by ๐ two. ๐ one is given as 0.025 meters,
while ๐ two is 0.015 meters. If we were to write a specific
equation then for the torque on our given current-carrying loop, we can replace the
cross-sectional area ๐ด of the loop with ๐ one multiplied by ๐ two. Another bit of information given to
us in our problem statement is that the current carried by our loop is 350
milliamperes, what weโll call ๐ผ. Along with this, weโre told that
the magnitude of the magnetic field is 0.12 tesla. Weโll call this ๐ต.

In this equation for torque, the
capital ๐, as we mentioned, refers to the number of turns in our loop. In this scenario, ๐ is equal to
one, and so we can drop it out of this equation. When it comes to the sin of this
angle ๐, we can note that ๐ is the angle between the external magnetic field and
the area vector of a rectangular loop. Based on the direction of current
in the loop, that area vector would point directly downward. Notice this means that the angle
between this area vector and the external field is 90 degrees. When we evaluate the sin of 90
degrees, we get one. And we can again drop out this
factor. It wonโt have an effect on the
final torque acting on the loop.

For our next step, we can
substitute these known values into our equation for torque. In place of ๐ต, we put 0.12
tesla. In place of the current ๐ผ, instead
of 350 milliamperes, we convert this to a number in amperes, 0.350. Note that weโve shifted the decimal
place three spots to the left to move from milliamperes into amperes. Then, ๐ one is 0.025 meters and ๐
two is 0.015 meters. In scientific notation, the exact
answer we get is 1.575 times 10 to the negative fifth newton-meters. In terms of units, note that one
newton is a tesla ampere meter.

So then we have an exact answer in
newton-meters. But our question asks us to give an
answer to the nearest micronewton-meter. A micronewton is one millionth of
one newton, meaning that if we wrote our answer as simply a decimal number, we could
convert it into micronewton-meters by moving the decimal place one, two, three,
four, five, six places to the right. We have then 15.75
micronewton-meters. Weโre almost there. We just need to round it to the
nearest whole number. The closest whole number to 15.75
is 16, which means that, to the nearest micronewton-meter, the torque acting on this
coil is 16 micronewton-meters.

Now, letโs look at part two of our
question.

Find the magnetic dipole moment of
the loop to the nearest micronewton-meter per tesla.

If we have a loop of
current-carrying wire, where the current has a magnitude ๐ผ and the area of that
loop is called ๐ด, then the magnitude of the magnetic dipole moment of the loop ๐
is equal to ๐ผ times ๐ด. To solve for ๐ in our case, weโll
use the fact that current in our rectangular loop is 350 milliamperes and that ๐
one and ๐ two, the perpendicular dimensions of our rectangular loop, are given.

Substituting these values into our
equation, notice that the units we have are milliamperes and meters times meters, or
meters squared. We want our final answer though in
units of micronewton-meters per tesla. Recalling that one newton equals a
tesla ampere meter, we can rearrange this to see that an ampere equals a newton per
tesla meter. All this means we can replace the
ampere here in our equation by a newton per tesla meter. And note that this m in our
equation is not actually the unit of meter but is the prefix milli-. Really, what we have is
millinewtons per tesla meters.

And then recall that weโre
multiplying this unit by the units of meters squared. When we multiply all these numbers
together, we get a numerical value of 0.13125, where the units are
millinewton-meters per tesla. Those arenโt exactly the units of
our final answer though. We want them to be in
micronewton-meters per tesla. To convert between these two, we
can recall that one millinewton equals 1000 micronewtons. This means that to change our units
to micronewton-meters per tesla from millinewton-meters per tesla, weโll need to
multiply this number here by 1000. Doing that gives us a result of
131.25 micronewton-meters per tesla.

Lastly, weโll report our final,
final answer to the nearest micronewton-meter per tesla. Rounding our answer to the nearest
whole number, we report that the magnetic dipole moment of the loop to the nearest
micronewton-meter per tesla is 131.