### Video Transcript

The diagram shows a rectangular
conducting coil with three turns that is in a magnetic field. There is a current of 8.5 amperes
in the coil. The sides of the loop parallel to
line π one are parallel to the magnetic field, and the sides of the loop parallel
to line π two are perpendicular to the magnetic field. The length of π one equals 0.035
meters, and the length of π two equals 0.025 meters. The torque on the loop is 15
millinewton-meters. Find the magnitude of the magnetic
field to the nearest millitesla.

In our diagram, we see our
rectangular conducting coil with three turns or loops in it. This coil is positioned between the
poles of a permanent magnet. For this reason, the coil is
exposed to a constant magnetic field that weβll call π΅. In this example, we want to solve
for the magnitude of that field to the nearest millitesla. To begin doing this, letβs clear
some space on screen and recall that the general equation for the torque acting on a
current-carrying loop in a magnetic field is given by this expression. Here, the magnitude of that
magnetic field is π΅, the current in the conducting loop is πΌ, the cross-sectional
area of the loop is π΄, and π represents the number of turns in the conducting
coil. Along with all this, we see that
thereβs an angle of π involved in this expression.

To understand π, imagine that we
have a rotating coil and a magnetic field, and we draw a vector thatβs perpendicular
to the plane of the rotating coil. The angle between that vector and
the external magnetic field, this angle here, that is π. We can see then that π will change
as our rotating coil turns.

If we look at our given scenario,
though, we see that in this orientation a vector that is perpendicular or normal to
the plane of our conducting coil looks like this. We see that the angle between this
vector and the external magnetic field is 90 degrees. As we look to apply this general
equation then, in our case, we can write that the torque π equals π΅ times πΌ times
π΄ times π times the sin of 90 degrees. The sin of 90 degrees equals
one. So, we can effectively leave out
this factor from our expression.

Letβs recall that itβs not the
torque we want to solve for, but rather the magnetic field magnitude. To do that, we can divide both
sides of the equation by πΌ times π΄ times π, canceling those factors on the
right. The magnetic field strength π΅
equals π divided by πΌ times π΄ times π. At this point, we can recall the
information given to us in our problem statement. We were told for one thing that at
this moment in time the torque acting on our conducting coil, weβll call it π, is
equal to 15 millinewton-meters. As we work with this term, weβll
just need to be careful to recall that this βmβ indicates milli-, the prefix meaning
one thousandth, while this βmβ indicates the unit meters.

We were also told the current,
weβll call it πΌ, that exists in our conducting coil. Itβs 8.5 amperes. Regarding the cross-sectional area
π΄ of our coil, we werenβt given that value, but we were told the dimensions, π one
and π two, of this rectangular coil. We can multiply those together, and
that will equal the area π΄. Lastly, our coil has more than one
loop to it. Weβll call that number capital
π. And we were told that itβs equal to
three. We can even see these three turns
in the coil in our diagram. Knowing all this, weβre now ready
to substitute into our equation and begin solving for π΅.

Note that as we do this, for the
variable π΄, we put in π one multiplied by π two. The product of those two values
will give us the cross-sectional area π΄. Thanks to the prefix milli- in the
numerator of our fraction, if we calculate this expression as itβs written, weβll
get an answer in units of millitesla; thatβs just what we want. And when we calculate this answer,
we get a result of 672.268 and so on millitesla. And now we recall that we want to
round our result to the nearest millitesla. Since the first decimal place digit
is less than five, our final answer is 672 millitesla. This is the magnitude of the
magnetic field surrounding our current-carrying conductor.