A student uses a bar magnet to lift some steel paperclips off a table. Figure one shows the magnet lifting the paperclips off the table. Explain how the bar magnet lifts a steel paperclip off the table.
Looking at Figure one, we see an image of something you may have experienced
before. If you start out with a group of paperclips sitting in a pile on a table and then
bring a bar magnet close to the pile without touching them, the paperclips actually
will do what we’re seeing here in Figure one. They will be drawn to the end of the bar magnet and attach in a chain like this,
hanging down from the magnet to the table.
Knowing that this does happen, we want to explain just how it is that this
happens. We know from experience that most materials don’t act this way. So what is it about the bar magnet and the steel paperclips that’s different?
Part of the explanation for all this comes from the fact that these paperclips are
made of steel. That means that this paperclip made of steel is in a class of materials that change
or respond when they’re in a magnetic field. Most materials of course aren’t like this. If you bring a bar magnet near to a leaf or to a book or to a cotton shirt, they
won’t change in response to the magnetic field from the bar magnet. But objects made of steel — like this paperclip — do.
Let’s consider how this happens. It starts with the fact that a bar magnet like the one we have in Figure one creates
a magnetic field around it. That field is invisible to our eye. But if we were to draw in what it might look like, if we could see it, it will look
something like this. The magnetic field lines — that’s what each one of these is called — start out at the
north pole of the magnet and move to the south pole.
Now, here’s what’s interesting about this. If we take our paperclip and put it in the field created by this magnet, not touching
the bar magnet, but just in the field it creates, then our paperclip like we
mentioned will be changed by this magnetic field. It will line up so that it is oriented along the magnetic field lines.
When this occurs, the paperclip itself becomes a magnet, not a permanent one like the
bar magnet, but a magnet nonetheless whilst in this field. Knowing that with magnets, north poles attract south poles and south poles attract
north poles, what do you think are the poles of this paperclip that’s in the
magnetic field we’ve drawn?
Well, if we start at the north pole of the bar magnet and follow the field lines
until they reach one of the ends of the paperclip, we know that because opposite
poles attract one another, that end of the paperclip will be its south pole. And likewise, if we start at the south pole of the permanent bar magnet and move
backwards along one of the field lines until we reach the paperclip, then that end
of the paperclip will be its north pole.
So the north end of the paperclip is closest to the south end of the bar magnet and
the south end of the paperclip is closest to the north end of the bar magnet. The paperclip that we’ve drawn is coming in from the side of the bar magnet. But in our scenario, in Figure one, we have a pile of paperclips right below it. A paperclip that’s lifted this way will end up looking like we see the paperclips in
One really fascinating thing about this effect is that the magnetic attraction
between the paperclip and the bar magnet is stronger than the force of gravity on
the paperclip; that’s why it rises up. Let’s put some of what we’ve described here into words as part of our answer.
The first step we mentioned in this process is that the magnetic field of the
permanent magnet makes the steel paperclips themselves magnetised; it changes
them. Next, we said that the field of the temporary magnet, that is the paperclip that’s
changed by the magnetic field of the bar magnet, aligns to the field that created
This is what it meant when we said that the north pole of our temporary magnet, the
paperclip, faces the south pole of our bar magnet. And it’s the same on the other side too. The relative south pole of our paperclip is aligned towards the north pole of our bar
Writing all that out, we can say that this means that the end of the paperclip
closest to the south pole of the magnet becomes the north pole of the paperclip. We say “becomes” here because our paperclip only turns into a magnet in the presence
of the bar magnet’s magnetic field. Without that field, the paperclip is unmagnetised and continues to sit in a pile on
Then, finally, because opposite poles attract and because the north pole of the
paperclip is facing the south pole of the bar magnet, the paperclip is lifted off
the table against the force of gravity. This is our explanation for just how it is that this physical phenomenon occurs.
In this example, we talked about how an object made of steel is affected by an
external magnetic field. Next, let’s consider what other types or materials are affected in a similar way.
Iron can be used to make permanent magnets. Which other materials can be used to make permanent magnets? Tick two boxes. Copper, magnesium, nickel, cobalt.
This question is interesting because it tells us that certain materials which do not
start off as magnets can become permanent magnets. We’re told that iron is a material like this and that there are two more in this list
for us to identify. Considering a large chunk of iron, let’s consider just how it would be that this
chunk could become a permanent magnet.
The idea is that iron and materials likes it are made up of what are called
domains. These domains are each like a little region within the larger chunk of material
itself. And each region in a material such as iron or one similar has its own overall
magnetic field. And the direction of the magnetic field can be represented by an arrow.
If we look at the directions of the magnetic fields in these domains of our chunk of
iron, we see that they point in pretty much every single direction: some point up,
some point down, some point left, some point right. If we consider the overall magnetic field of this entire chunk of iron, we can’t
really tell what that is because the magnetic fields of the domains point in all
these different directions.
Remember though we’ve been told that materials made of iron can become permanent
magnets. Here’s how that happens. Say we take our chunk of iron and we put it within a magnetic field. We’ve called that field capital 𝐵. This external magnetic field will change the domains of this chunk of iron; it won’t
change where their boundaries are, but it will change the direction of the magnetic
field in each domain.
Thanks to the magnetic field, look what happens to the magnetic fields in each of the
domains. They line up with that external field. If we were to then turn off the external magnetic field so that it’s no longer
influencing the domains of this chunk of iron, those domain fields would stay in
largely the same direction.
Over time, the magnetic fields of these domains would drift out of alignment. But for quite some time, they would be relatively in line. And that’s why we say that iron can be made into a permanent magnet. The property that iron has of having domains each with their magnetic field and
having these magnetic fields respond to an external magnetic field has a name for
That name is ferromagnetism, where the prefix ferro refers to iron and magnetism of
course refers to magnetism. And it turns out that iron isn’t the only ferromagnetic material there is. There are other elements that behave this way too that are ferromagnetic.
We can find out whether a given element is ferromagnetic by looking up its properties
in a table. When we do this for copper, magnesium, nickel, and cobalt, we see that it’s the last
two elements listed here, nickel and cobalt, which both themselves are
ferromagnetic. That means that these elements like iron can be used to make permanent magnets.
Next, let’s consider a question not about magnetic poles, but about electric poles,
that is, electric charges.
Figure two shows two parallel, charged plates. The charged plates create a uniform electric field between them. Use an expression from the box to complete the sentence: decreases, stays the same,
increases. As the amount of charge on each metal plate is increased, the potential difference
between the positively charged plate and the negatively charged plate blank.
In this question then, we want to work off of Figure two which shows a positively
charged plate opposite and negatively charged plate. Based on what’s going on there, we want to choose from this list of possible answers,
decreases, stays the same, or increases, to fill in the blank in this sentence.
This sentence that we want to complete talks about the amount of charge that’s on
each of our two metal plates. We know from Figure two that our top plate has a positive charge and our bottom plate
has a negative charge. And if we were to sketch those charges in on our plates, they might look something
Not only do we have two oppositely charged plates, but we’re told that a uniform
electric field exists between these plates. That’s actually to be expected because electric charges of any sign whether positive
or negative create an electric field around them. We see those electric field lines drawn in between our positive and negative plates
and the arrows represent which direction the field lines point, from positive to
The question is what would happen if we increased the amount of charge on each plate;
that is, what would happen if we put more positive charge on the positive plate and
more negative charge on the negative plate? Well, for one thing the electric field in between these two plates would change. Just as the charge increased, so will the strength of that electric field.
We could represent that change if we wanted by drawing more electric field lines to
indicate that the field is getting stronger. But it’s not the electric field strength that we’re interested in directly, rather
it’s the potential difference between the positive and negatively charged
Here’s what we know so far. We’ve increased the charge on each plate and that increase has led to an increase in
the strength of the electric field between the plates. The question is how does our increased electric field relate to the potential
difference between the plates.
In cases like this, where we have oppositely charged parallel plates facing one
another, there is a helpful relationship between electric field and potential
difference that we can recall.
If we represent the potential difference between the plates as capital 𝑉, then that
potential difference it turns out depends on two factors: one is the distance
between the two plates — we can call that distance 𝑑 — and the other is the
electric field that’s in between these oppositely charged plates.
If we take the product of that electric field and the distance between the two
plates, that product is equal to the potential difference between them. Now, let’s apply this to our situation.
We know that in this instance the distance between our plates isn’t changing; that’s
a constant value. But what is changing? Well, the electric field is changing. And it’s increasing because we’re putting more charge on the positively and
negatively charged plates.
So if the distance is staying the same and the electric field is increasing, then
what must be happening to the potential difference 𝑉? Since the right-hand side of this equation is increasing, that means the left-hand
side must be increasing as well. Potential difference must be going up.
This tells us which expression from the box to choose to complete the sentence. As the amount of charge on each metal plate is increased, the potential difference
between the positively charged plate and the negatively charged plate increases. That’s how this change in charge affects potential difference.
Finally, let’s work on a calculation involving potential difference.
When the potential difference between two charged plates is 50 kilovolts, a spark
jumps between the two plates. The spark transfers 0.00004 coulombs of charge between the plates. The equation which relates the energy transferred by a charge that moves across an
electric potential difference, the size of the charge, and the potential difference
is energy transferred equals charge times potential difference. Calculate the energy transferred by the spark.
In this example, we have a series of information given to us and then a way to
combine that information to give us our desired result: the energy transferred by
this spark that moves between two charged plates. To calculate this amount of energy, let’s start out by using symbols to represent the
information we’re given as well as this energy transfer equation.
Beginning with the equation, let’s use capital 𝐸 to represent the energy
transferred. We’ll use capital 𝑄 to represent charge and we’ll use capital 𝑉 to represent
potential difference. Off to the side then, we can summarize this equation as 𝐸 equals 𝑄 times 𝑉. And we’ll now know what that means.
The charge transferred between plates multiplied by the potential difference between
them is equal to the energy that’s transferred between the plates. If we looked to our question statement, we see we’re given information about that
potential difference 𝑉 as well as the charge 𝑄. We can write those in shorthand too.
Clearing some space, we can say that 𝑉, the potential difference, is equal to 50
kilovolts and that 𝑄, the charge transferred between the plates, is equal to
0.00004 coulombs. We want to combine 𝑉 and 𝑄 according to this equation to solve for 𝐸. But before we do it, there’s a change that we’ll need to make.
That change involves the units of our potential difference 𝑉. Notice that those units are given in kilovolts that is thousands of volts. When we use potential difference in our equation though, we want its units to simply
be units of volts. So before we calculate 𝐸, let’s convert our potential difference from units of
kilovolts to units of volts.
Here’s how we’ll do that. We’ll recall the conversion between kilovolts and volts. If we take 1000 volts, that’s equal to one kilovolt. In fact, that’s what the prefix kilo refers to, 1000. So this means that our voltage given as 50 kilovolts is actually 50 times 1000
We can write it this way. We can say that since a kilovolt is equal to 1000 volts, then we can replace
kilovolts with that expression, 1000 volts. And we’ll multiply that by 50 to get our potential difference 𝑉 in volts. 50 times 1000 is equal to 50000. So that’s the number of volts of our potential difference let’s use this updated
value in place of our previous value for 𝑉.
And realise that both values are the same; they’re just expressed in different
units. We’re now ready to solve for the energy transferred by multiplying 𝑄 times 𝑉. We plug in the values for 𝑄 and then for 𝑉 and multiply them together. And when we do, we find a result of 20 joules. That’s the amount of energy transferred by the spark.