### 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?

Let’s start by recalling that
moving charges in a magnetic field experience a magnetic force represented by 𝐹 sub
𝐵. So, because this conducting rod is
connected as part of a circuit, and since it’s within a magnetic field, a magnetic
force will act on the rod and cause it to accelerate. Our job is to find the magnitude of
that acceleration.

To start, we know that force is
equal to mass times acceleration or that acceleration equals force divided by
mass. And we already know the mass of the
rod, so we just need to calculate the magnetic force acting on it.

To do this, let’s recall that for a
straight conductor of length 𝐿 that’s carrying a current 𝐼 and moving
perpendicular to the direction of a magnetic field of strength 𝐵, the magnetic
force on the conductor is given by 𝐵 times 𝐼 times 𝐿. We know values for 𝐵 and 𝐿, but
we don’t know the value of the current in the rod, 𝐼. So let’s recall that by Ohm’s law,
current in a circuit equals the potential difference across the circuit divided by
its resistance. We can go ahead and substitute this
expression into the formula for the magnetic force. So it becomes 𝐵 times potential
difference over resistance times 𝐿.

Finally, we have an expression for
𝐹 sub 𝐵 in terms of values that we know. But remember, we’re really wanting
to solve for acceleration. So let’s divide this entire
expression by the mass of the rod. Doing this and moving resistance to
the denominator, we have our equation for the rod’s acceleration due to the magnetic
force. We’re getting really close. But before we go any further, we
should make sure that all the values for the terms on the right-hand side are
expressed in their appropriate base SI or SI-derived units.

Let’s start with the strength of
the magnetic field, 𝐵, which equals 125 milliteslas. This should be written in plain
teslas. So 𝐵 equals 0.125 teslas. Next, the potential difference
equals 4.5 volts. So it’s good to go, along with
resistance, which equals 2.5 ohms. Moving on, the length of the rod 𝐿
is given as 15 centimeters. So let’s rewrite this as 0.15
meters. And last, we know that the mass of
the rod is 750 grams or 0.750 kilograms.

We’re now ready to plug these
values in. And because they’re all expressed
in their appropriate units, we know that we’ll end up with an acceleration value in
proper units of meters per second squared. Finally, grabbing a calculator, we
get a result of 0.045 meters per second squared. This is the magnitude of the
acceleration of the rod.