### Video Transcript

The circuit shown in the diagram
contains a 4.5-volt battery attached to smooth conducting rails. The ends of the conducting rails
are attached to a 15-centimeter-long conducting rod with a resistance of 2.5 ohms
and a mass of 750 grams. The circuit is within a
125-millitesla uniform magnetic field. What is the magnitude of the
acceleration of the rod?

In our diagram, we see this
conducting rod resting at either end along conducting rails. These rails are electrically
connected to a 4.5-volt battery. Even when the rod is stationary,
electrical charge can flow through the circuit. Because charge is moving through
the conducting rod and that rod exists in a uniform magnetic field, it experiences a
magnetic force. The magnitude of that force is
equal to the magnitude of the magnetic field strength ๐ต multiplied by the strength
of the current ๐ผ times the length of the conducting rod ๐ฟ.

Our question though doesnโt ask
about the force on the conducting rod but rather its acceleration. Net force and acceleration are
connected though through Newtonโs second law of motion that ๐น is equal to ๐ times
๐. Here, ๐น is the force on our
conducting rod, ๐ is the rodโs mass, and ๐ is its acceleration. Combining these equations, we can
write that ๐ times ๐ equals ๐ต times ๐ผ times ๐ฟ. Then, dividing both sides by the
mass of the rod ๐ so that that mass cancels on the left, we have an expression for
the acceleration of this conducting rod.

Note that weโre given the magnetic
field strength ๐ต. Thatโs 125 milliteslas. Weโre also told the length of the
rod ๐ฟ, thatโs 15 centimeters, and the rodโs mass ๐, 750 grams. The one thing we donโt know is the
current ๐ผ that exists in the rod. But notice that weโre told that the
battery supplies 4.5 volts of potential difference and the resistance of the rod is
2.5 ohms. If we assume that this rod follows
Ohmโs law, then ๐ is equal to ๐ผ times ๐
, which means, rearranging slightly, that
๐ผ is equal to ๐ divided by ๐
.

We can make that substitution in
our equation for the acceleration ๐. In place of the current ๐ผ, we have
๐ divided by ๐
. So now ๐ equals ๐ต times ๐ times
๐ฟ divided by ๐
times ๐. As weโve seen, we do know ๐,
thatโs 4.5 volts, and ๐
. Thatโs 2.5 ohms. So we now know everything on the
right-hand side of this expression and are ready to substitute in. What we have is a magnetic field of
125 milliteslas, a voltage of 4.5 volts, a length of 15 centimeters, a resistance of
2.5 ohms, and a mass of 750 grams.

Before we calculate ๐, weโll want
all the units in this expression to be on the same footing. To do that, weโll convert
milliteslas to teslas, centimeters to meters, and grams to kilograms. Clearing some space at the top of
our screen, we note that 1000 millitesla equals one tesla. 100 centimeters is a meter. And 1000 grams is a kilogram. Our conversion for milliteslas to
teslas means weโll take the decimal point in this value and will move it one, two,
three spots to the left. This is the same as dividing by
1000. 125 milliteslas is 0.125
teslas.

In a similar way, to convert
centimeters to meters, weโll take the decimal point in this value, and weโll move it
two spots to the left. This is like dividing by 100. 15 centimeters is 0.15 meters. Lastly, weโll convert our mass in
grams to kilograms by moving the decimal point three spots to the left. 750 grams is 0.750 kilograms.

Weโre now ready to calculate
๐. And weโll get an answer when we do
in units of meters per second squared. We find that ๐ is exactly 0.045
meters per second squared. This is the magnitude of the
conducting rodโs acceleration.

Letโs look now at part two of our
question.

What is the initial rate at which
the potential difference across the rod changes due to the emf induced across it
because of its motion in the magnetic field? Give your answer in scientific
notation to one decimal place.

In this part of our question, we
realize that not only is there a potential difference across the rod because itโs
part of the circuit with a 4.5-volt battery, but it also experiences a potential
difference because of emf generated as it moves. As we saw in part one of our
question, the rod experiences an acceleration even if it starts from rest due to the
fact that itโs part of the circuit. That acceleration is to the right,
moving the rod through, as we see, a uniform magnetic field. For a rod of length ๐ฟ moving
perpendicularly through a uniform magnetic field of strength ๐ต at a speed ๐, the
emf, or potential difference induced across the rod, is ๐ times ๐ต times ๐ฟ.

In this part of our question, we
want to solve for the initial rate at which the potential difference across the rod
changes. Symbolically, we want to solve for
a change in potential difference, weโll represent that as ฮ๐ธ, over some amount of
time, represented by ฮ๐ก. Notice then that weโve taken ๐,
representing potential difference, put a ฮ sign in front of it to show that it is
changing, and then divided this quantity by ฮ๐ก. We can do the same thing then to
the other side of the equation. We put a ฮ, or a change sign, in
front of ๐ times ๐ต times ๐ฟ, and we divide all this by ฮ๐ก.

Letโs note here that of these three
quantities, ๐ and ๐ต and ๐ฟ, only one of them changes in time. Thatโs the velocity ๐. We know this because as we saw from
part one of our question, our rod is accelerating. That means we can drop our
parentheses and write our numerator as ฮ๐, a change in velocity, times ๐ต times
๐ฟ.

Looking at this expression, we can
recall something about ฮ๐ divided by ฮ๐ก. Acceleration ๐ is defined as a
change in velocity ฮ๐ over a change in time ฮ๐ก. Therefore, we can rewrite this
expression as ๐, the acceleration of the conducting rod, times ๐ต, the magnetic
field magnitude, times ๐ฟ, the length of the rod. Letโs recall from earlier that ๐
is 0.045 meters per second squared, ๐ต is 0.125 teslas, and ๐ฟ is 0.15 meters. Substituting these values into our
equation, what weโre going to calculate is a change in potential difference that is
in volts divided by a change in time in seconds. Writing our answer in scientific
notation to one decimal place, we get 8.4 times 10 to the negative four volts per
second.

Notice that this answer depends on
the acceleration of the rod but it doesnโt depend on its velocity. At this initial moment, the rod
could be in motion or it could be at rest. And this change in potential
difference over time will be the same. As we see, it only depends on the
rodโs acceleration, the strength of the magnetic field, and the rodโs length.