- Hi everyone. My name is Anika.
Thank you for joining today.
Today, we're going to walk through
a sample chem phys passage,
and it's going to be all about electrostatics and magnetism.
Everyone's favorite physics topic.
So let's just get right into it.
Normally, I like to recommend spending
about like eight ish minutes on a passage like this,
but for today,
we're just going to walk through it all out.
So it will take a little bit longer.
So let's just get into it.
I'm going to go ahead and read and highlight
what I think is important.
And then we'll talk through
some of the important figures in this passage.
So while conducting experiments
on the tissue of a newly identified type of tumor,
researchers discovered a previously unidentified molecule,
molecule alpha.
Preliminary experiments with molecule alpha
indicate that it is charged.
And researcher's first plan is to calculate its charge
by placing a sample between
two of a large parallel plate capacitor.
After charging the capacitor,
using a nine volt battery,
the researchers measure a one times 10 to the negative 12
Newton force pulling molecule alpha
towards the negatively charged plate.
Okay, let's take a look at the figure
that they give us to sort of
make sense of the words that we just read.
So here we know that we're putting the sample
in between something.
So here we have our sample that adds up,
we're putting it between the two plates
of a parallel plate capacitor.
So this looks like it's our capacitor.
We know that capacitors have electric fields through them.
And so we see the electric field.
We know that the electric field is pointing downwards
just by the way the field lines look.
We also have a distance for our electric field,
which is great.
And we know that this is an electric field
because we have a QV right here
in this sort of circuit situation. Okay.
So that's,
I think, all we need to take away from that figure.
Let's keep reading.
Researchers perform a second experiment
to determine the mass of the molecule using a cyclotron.
The cyclotron contains a tunnel with an electric field
and magnetic field pointing in perpendicular directions
across its width.
A charged particle is ejected through the tunnel
at a characteristic velocity.
The electric field and magnetic field are oriented such
that their exerted forces on the molecule sum to zero.
That seems important.
The particle then emerges from the end of the tunnel
into a space where the magnetic field remains,
but the electric field is turned off.
Remaining under the force of the magnetic field,
the particle undergoes circular motion
and travels to distance 2R from the opening of the tunnel.
Okay. That was a lot of words.
Let's take a look at this picture of the cyclotron.
So it looks like we're kind of separating the cyclotron
into two parts. There's a first part here,
which is characterized by this tunnel
that we've been hearing about.
And there's the second part here.
So if we think about this tunnel,
the passage tells us that the electric field
and magnetic field are oriented
so their forces are perpendicular
to the width of the tunnel.
And we also know that their opposite forces sum to zero.
So that tells us that they must be pointing
in opposite directions.
We can see that our electric field lines run downwards.
So therefore, our magnetic field lines must point up,
our magnetic force must point upwards
if our electric force is pointing downwards.
We could, if the passage didn't tell us
that that force was zero
and we didn't know what direction it was,
we could also think about using the right hand rule.
And the way I like to do that is to point my hand
in the direction of the velocity,
curl my fingers in the direction of the B field,
the magnetic field, and whichever direction
my thumb is pointing is the magnetic force.
So for this problem,
our velocity is to the right.
So I'm going to point my hand this way.
You can see X's for our B field.
So that means I'm going to curl my fingers into the page.
And my thumb is just naturally pointing upward
so our magnetic force will also be pointing upwards,
and that adds up;
the electric force downward magnetic force upwards.
So that's the first part when we're in the tunnel.
Once the sample leaves the tunnel,
we know that we enter this phase
where the electric field is supposedly turned off,
which means that the only field we have here
is the magnetic field, which also means
that there's no electric force.
There's only magnetic force at this point.
We also see our molecule take
sort of a circular trajectory,
and that seems like it will be important for later.
Okay.
That's I think pretty much all I want to talk about
before we get into the questions.
And that's sort of a good understanding of like
the figure, and just sort of extrapolating
the text onto what's going on in the experimental setup
before you approach the questions.
Okay. Let's start with number one.
What is the magnitude between the,
or what is the magnitude of the electric field
in between the plates of the capacitor?
Assume this experiment takes place in a vacuum.
Okay. So for this kind of problem,
it would probably be really helpful
to know the Equation for the electric field,
for an electric field.
If we didn't however,
remember the Equation and we're taking our test,
one thing we could do is take a look at the units
and see if we could reverse engineer the problem.
So we see that the numerator of our units
is some kind of voltage,
and the denominator is some kind of distance.
This V stands for voltage, not velocity.
We know that because it's a capital V.
So if we took it, take a look at the values
they give us in our problem,
we see that we're using a nine volt battery.
That sounds like a good number for our voltage.
So let's put that on the numerator.
And our distance we already talked about
was the distance between the two plates,
which is 0.01 meters.
So if we do nine divided by 0.01 meters,
that's also Equal to nine divided by one times 10
to the negative two,
which is also Equal to nine times 10 to the positive two.
And so let me just write that out.
Nine times 10 to the positive two
is Equal to 900 volts per meter,
and that corresponds to answer choice D.
Okay, great.
The Equation that would have probably been helpful
if we just memorized it was E Equals V over d
but as you can see,
we were able to get the correct answer
without having that memorized.
All right. What is a charge of molecule alpha?
Okay.
So it looks like we have some positive choices
and some negative choices.
So I think the first step would be to figure out
what sign our charge is going to be.
So if we remember from any time we learned about magnets,
key thing is that opposites will be attracting
and like charges will be repelling.
So in this case, our passage tells us that there's a force
that's pulling the molecule
towards a negatively charged plate.
If something is just naturally being pulled
towards a negative if it's attracted to the negative,
it must be positive.
If it was negative, it would be repelled by the negative.
So we're going to go ahead and eliminate A and B
because we know our charge has to be positive.
Unfortunately,
we still have to calculate the value of it,
but it doesn't hurt to have eliminated it down to 50%.
Okay.
So what is a charge on molecule alpha?
Here's where it's going to be really important
to remember some formulas
and manipulations of electric fields.
And so the formula that's going to be really useful
for this problem is F Equals E Q.
And this is a derivation of Coulomb's Law
if any of you guys remember that,
and this E is basically referring to the electric field,
this Q is referring to our molecule alpha's charge.
And the force is essentially the force on molecule alpha
that's generated by this electric field.
So we have our E,
we actually just calculated that in the previous problem,
we have our F, our passage tells us is
one times 10 to the negative 12.
Now let's calculate for Q.
So Q Equals F over E,
which is going to be one times 10 to the negative 12 Newtons
divided by E, which is going to be 900 volts per meter,
which is the same as nine times 10 to the two
volts per meter.
I like to separate out my like new number
versus scientific notation part of,
or exponent part of the math.
And so what I'm going to do is one over nine times,
let's figure out the 10 to the negative 12
divided by 10 to the two is the same
as 10 to the negative 14,
which is Equal to 0.11 times 10 to the negative 14,
which is Equal to 1.1 times 10 to the negative 15.
Okay. Lots of calculations,
and I have really bad handwriting,
but that is our correct answer.
1.1 times 10 to the negative 15 Coulombs.
Okay. Number three,
at what velocity should a sample of 1010 molecules
be fired into the tunnel of the cyclotron
so that it will make it to the end of the tunnel?
Okay. So we can look at the picture again,
and we know that we're talking about the tunnel phase
as specified by the question.
We talked about how here we have an electric force
pointing downwards and a magnetic force pointing upwards.
The passage tells us that the,
the forces have to sum to zero.
And let's think about why that might make sense.
If one was greater, let's say, for example,
our electric force was way greater than our magnetic force.
That would just automatically incentivize anything
we place into the tunnel to shoot straight downwards, right?
If our electric force is overpowering our magnetic force,
then the molecule wouldn't even want to go horizontally.
It would just want to go straight down
in the direction of the net force.
But if our net forces sum up to be zero,
then the molecule has no reason to go up or down.
It can go straight across
with whatever characteristic velocity
we shoot through the tunnel with.
So we know that the basic supposition
of what's happening in our experimental setup
is that the two forces have to sum,
or the two forces have to be Equal to each other
so that their net force is zero
since they're in opposite directions.
So let's set the forces Equal to each other.
We talked about already how in the previous question,
our electric force is going to be Equal to Eq.
Now, all we have to do is set that Equal
to our magnetic force
and remember what an Equation for magnetic force could be.
And that Equation,
which you're going to want to remember
for your MCAT is qVB.
B here is referring to magnetic field.
If we're solving for V, velocity,
we can go ahead and cancel out the q's.
And we're left with v Equals E over B meters per second.
And that is A.
Okay, you'll notice that I didn't even bother
with the fact that we're talking about 1010 molecules.
And the reason for that is because
I decided to look at what was happening
at a one molecule scale.
So for example, Eq is the electric force on one molecule,
qVB is also the magnetic force on just one molecule.
By doing that,
I was able to cancel everything out
and look at what the velocity of one molecule would be.
Let's think about what this would mean for 1010 molecules.
The electric force on 1010 molecules
would just be a 1010 times Eq and the electric force on,
or the magnetic force on 1010 molecules
would just be a 1010 times qVB.
Regardless of how many molecules we're looking at,
that number will cancel out.
So it doesn't really matter how many molecules we have.
We're always gonna end up with the same net velocity.
All right, let's look at number four.
What is the acceleration experienced by the sample
once it enters a space outside the tunnel?
So let's take a look at the picture.
If we're outside the tunnel,
we are now in this second phase right here,
and we know the passage told us that
once we've gone outside the tunnel,
our electric field shuts off
and we only have a magnetic field.
So that means the only force that our molecule
is experiencing is a magnetic force.
Also, the passage tells us
that the molecule undergoes circular motion.
And we can kind of see that
just by looking at the picture of these dashed lines.
The molecule takes a sort of a circular trajectory.
And so that is kind of an indication
that we're not going to be thinking
about normal acceleration here.
We want to be thinking about centripetal acceleration.
And one thing we'll want to remember is that
the Equation for centripetal acceleration
is V squared over R. Okay?
If it was normal acceleration,
we could use F equals ma or some derivation of that.
But in this case here,
we have to use centripetal acceleration.
We already solved for V,
which is very nice in our previous question,
which we decided was going to be E over B.
We'll square that and divide by R.
And we end up with E squared over B squared R
which corresponds to A.
A equals E squared over B squared R.
Okay, perfect.
Lastly, let's take a look at number five.
In terms of E B, q and r what is the mass of the sample?
Okay. So we're looking for a mass.
One really quick trick
for problems like these is just to take
a quick look at the answer choices
and make sure all of them are in terms of E, B, q and r.
If any of these answer choices has a variable,
which is not a fixed quantity, a variable,
that's not one of these letters,
we can go ahead and eliminate it
because it's not in its final form.
So in this case, if we look at B,
we have an A in our answer, that's an acceleration.
Our question is not asking us to get our answer
in terms of A so we can go ahead and eliminate that.
Similarly, we can also go ahead and eliminate D.
So that's a quick way to just eliminate some things.
Unfortunately, we still have to solve
the entire problem all out,
but we were able to eliminate half the choices.
Okay.
So what is the mass of the sample?
Here is where we want to use
our most fundamental ever physics Equation,
which is F Equals ma.
And if you go into your MCAT knowing just one equation,
I hope it's this one,
because this is a very, very versatile Equation.
And it's going to come in handy here.
So we found our A in our previous problem,
we decided that was going to be
E squared over B squared times R.
We're looking for m. What about F?
And so remember, we're in this phase of the cyclotron
where the electric field is turned off.
All we have is magnetic field,
which means that our force is just purely magnetic force.
And we talked about how in a previous question,
we talked about how magnetic force is equal to qVB.
So solving for m, we can use F over a,
and let's just do some algebraic manipulation.
F is going to be qVB, as I just said.
A is going to be V squared over R,
one of these Vs are going to cancel out
and we're left with m Equals to be our qBr over V.
Our V is from question two, E over B.
So this is going to be qBR over E over B.
Bear with me, I know my handwriting is really terrible,
and that is going to Equal qB squared r over E.
Okay. Lots of algebra.
And I was writing really slow on this board,
but that corresponds to,
let's make sure Q B squared R over E,
Q B squared R over E.
Perfect. Okay.
That corresponds to letter choice C.
Okay, great.
So yeah, that wraps up this passage.
Please feel free to email me or comment
if you have any questions.
(upbeat music)