Video Transcript
In this video, our topic is the
magnetic field due to a current in a solenoid. We’ll see how this field is
created, what it looks like, and we’ll also learn how to calculate the field
strength within the solenoid.
To begin, we can recall that any
current-carrying wire like this one here will produce a magnetic field around
itself. And further, if we take the ends of
this wire and we connect them together to form a loop, then that current-carrying
circular loop also creates a magnetic field within as well as outside of itself. Now, if we take this loop of wire
and we angle it a bit, so we’re looking at it almost side on, what happens — even
though it’s a little hard to see with all the magnetic field lines — is that inside
this loop of current-carrying wire, a net magnetic field to the left is created.
This is the result of one circular
loop. But what if we had a number of
loops like this, all with current pointed in the same direction? If that were so, then each
individual loop would also contribute to this magnetic field pointing to the left
inside the loops. The net result is a fairly strong
and fairly uniform field pointed that way. Now, what if instead of being
separate, these loops were all joined together into one continuous wire? If this were so, we would find that
the magnetic field that’s created is essentially the same as that created by our
several loops. The name we give to a wire that’s
arranged this way is solenoid.
If we looked at a solenoid end on,
it might just look like this, like a circle. But seen from the side, we see all
the different turns or loops that are a part of this solenoid coil. When a solenoid carries current, as
this one does, it indeed creates a strong magnetic field within the turns of the
solenoid. We can represent that with these
multiple field lines. And it also creates a magnetic
field, albeit a weak one, outside the solenoid.
Now, considering this end-on view
of the solenoid here, let’s say that we got that view by putting our eye over at
this end. This means that if we could see the
magnetic field within the turns of the solenoid, it would look something like
this. Thanks to the total effect of all
the different turns in the solenoid, the strength of the magnetic field in this
cross section that we’re looking at right here is effectively constant. Whether we’re talking about the
magnetic field strength here or here or here in our cross section or at any other
point, that value is essentially the same.
There’s an equation to describe the
strength of the magnetic field inside the loops of the solenoid. If we call the strength of that
field 𝐵, then 𝐵 is equal to this constant, 𝜇 naught, which represents the
permeability of free space — essentially, how easily magnetizable free space or a
vacuum is — multiplied by 𝐼, the magnitude of the current in the solenoid, times
capital 𝑁, where capital 𝑁 represents the number of loops or turns the solenoid
has. And then all this is divided by the
length of the solenoid along its axis referred to as capital 𝐿.
In the solenoid we’ve sketched on
screen then, capital 𝑁 will be equal to one, two, three, four, five, six, seven,
since we have seven turns in our solenoid. And 𝐿, the solenoid’s length,
would be equal to this distance right here. Those are the values we would use
in this equation to calculate the strength of the magnetic field inside the turns of
the solenoid.
It’s worth mentioning that just as
the magnetic field magnitude is the same in any given cross section of the
solenoid’s interior, we’re also saying it’s the same in magnitude along the length
of the solenoid. So the magnetic field strength here
inside the turns of the solenoid is equal to the magnetic field strength here, say,
also within the solenoid core. Basically, there’s this volume,
we’ve shown it here in pink, and we’re saying that the magnetic field strength at
any point within that volume is given by this equation here.
Now, if we rearrange this equation
a bit so it reads 𝜇 naught times 𝑁 times 𝐼 over 𝐿, sometimes we’ll see this
equation written in an alternative form where capital 𝑁 over 𝐿 is replaced by a
lowercase 𝑛. When the equation is written this
way, we say that it’s written in terms of the turn density of the loops in the
solenoid. In other words, how many loops or
how many turns there are per unit length. So, the equation for magnetic field
can be written either way, and each form means the same thing. But this form, as we’ve seen, is a
bit more condensed and that it uses this factor called the turn density. When we use lowercase 𝑛, it means
that we don’t need to know specifically how many total turns a solenoid has or
exactly how long it is. We’re just interested in the ratio
capital 𝑁 to 𝐿.
So that’s how we calculate the
magnitude of the magnetic field within the loops of a solenoid. And as we saw, within the turns of
a solenoid, the direction of the field is always the same. The direction of the field will
depend on the way the current points in the solenoid, but it will always point along
the solenoid’s long axis, either to the left or to the right when we view a solenoid
side on. Notice though what happens when we
move outside these turns. In this case, if we follow a field
line, say this one, we see that it starts to loop back in on itself and eventually
forms a closed loop. By the way, this is true of
magnetic field lines in general.
So, on a diagram showing us the
magnetic field created by a solenoid, we can expect that whichever direction the
field within the loops of the solenoid points, in this case that direction is to the
left, the field lines outside the solenoid will point the opposite direction as they
turn back on themselves. But then, once these field lines
enter the solenoid loops, they do point in the same direction as all the other field
lines. Knowing all this, let’s get a bit
of practice with these ideas through an example.
A solenoid has a length of 3.2
centimeters and consists of 90 turns of wire. The wire carries a current of 1.2
amperes. Calculate the strength of the
magnetic field at the center of the solenoid. Give your answer in teslas
expressed in scientific notation to one decimal place. Use a value of four 𝜋 times 10 to
the negative seventh tesla meters per ampere from 𝜇 naught.
Okay, so in this exercise, we’re
working with a solenoid, which is a series of turns or loops of wire. Now, we’re told that our solenoid,
has 90 turns, and we can see there definitely aren’t 90 here, but we can just
pretend there are. So this solenoid has 90 turns, and
we’ll say that’s represented by the letter capital 𝑁. Along with this, we know that the
length of our solenoid, we’ll call it capital 𝐿, is 3.2 centimeters and also that a
current, we’ll call it 𝐼, of 1.2 amperes is carried by this wire.
Because of this current, the turns
in the solenoid will create a magnetic field around themselves. We want to calculate the strength
of that cumulative or overall field at the center of the solenoid. So, that would be about at this
point along its length. And if we were looking at the
solenoid end on through its circular cross section, we would find that point right
here. So it’s there that we want to solve
for the strength of the magnetic field, and we’ll call that field strength capital
𝐵.
At this point, we can recall a
mathematical relationship describing this magnetic field. The strength of the field at the
center of the solenoid is equal to 𝜇 naught, a constant called the permeability of
free space, multiplied by the total number of turns the solenoid has times the
current that exists in this wire all divided by the overall length of the solenoid
from end to end. In our problem statement, we’re
told what 𝑁, 𝐼, and 𝐿 are, and we’re also given the value to use for 𝜇
naught. So, we can substitute those given
values into our equation for 𝐵. So then, the value we use for 𝜇
naught is four 𝜋 times 10 to the negative seventh tesla meters per ampere. Capital 𝑁 is 90. The current 𝐼 is 1.2 amperes. And the length of the solenoid 𝐿
is 3.2 centimeters.
Before we calculate this magnetic
field strength, notice that the units of amperes, A, here cancel out in our
numerator and that also, in the numerator, we have these units of meters while in
the denominator we have a distance unit in centimeters. We’re going to give our answer in
teslas. We see that’s the unit right here
represented by capital T. And to do this, we’ll need our distance unit to cancel
out. To help that process along, let’s
convert the length of our solenoid from centimeters into meters. 100 centimeters is equal to one
meter. And therefore, 3.2 centimeters is
0.032 meters. And now, the distance units in our
numerator and denominator, written in meters, will indeed cancel out and we’re left
with units of teslas.
And when we calculate 𝐵, we get a
result in scientific notation to one decimal place of 4.2 times 10 to the negative
third teslas. This is the strength of the
magnetic field at the center of solenoid.
Let’s look now at another example
exercise.
A wire that carries a constant
current of 0.9 amperes is formed into a solenoid of length 310 millimeters. The strength of the magnetic field
at the center of the solenoid is measured to be 7.7 times 10 to the negative fourth
teslas. Calculate the number of turns used
to form the solenoid, giving your answer to the nearest whole number of turns. Use a value of four 𝜋 times 10 to
the negative seventh tesla meters per ampere for 𝜇 naught.
In this example then, we have a
solenoid that carries a constant current, we can call it capital 𝐼, of 0.9 amperes
and whose length is 310 millimeters. And we’ll call this length capital
𝐿. The solenoid is made of an unknown
number of turns. We’ll call that number capital 𝑁,
and that’s what we want to solve for. To help us do this along with the
current 𝐼 and the length 𝐿, we’re told the strength of the magnetic field at the
solenoid center. We can call that field 𝐵, and its
strength is given as 7.7 times 10 to the negative fourth teslas. In order to calculate capital 𝑁,
the total number of turns in the solenoid, we’ll want to recall how this variable is
related to the variables of magnetic field strength, current, and length.
The strength of the magnetic field
at the center of a solenoid is given by 𝜇 naught, this constant called the
permeability of free space, multiplied by the total number of turns in the solenoid
times the current that exists in it all divided by the length of the solenoid along
its axis. In our case, it’s not 𝐵 that we
want to solve for but the number of turns, 𝑁. So to do that, let’s multiply both
sides of the equation by 𝐿 over 𝜇 naught times 𝐼. Over on the right-hand side, this
means that 𝐿, 𝜇 naught, and 𝐼 all cancel out. And we find that the number of
turns in a solenoid is equal to its length times the magnetic field strength at its
center divided by 𝜇 naught times the current in the solenoid 𝐼.
When it comes to the factors on the
left-hand side of this expression, we’re given all four of them. We know 𝐿, 𝐵, and 𝐼. And we’re told to use a value of
four 𝜋 times 10 to the negative seventh tesla meters per ampere for 𝜇 naught. Substituting all these values in,
we find this expression for the number of turns 𝑁. Before we calculate this value
though, let’s convert this length of our solenoid, which is currently in units of
millimeters, to units of meters. To help us do that, we can recall
that 1000 millimeters equals one meter, which means that to convert 310 millimeters
to meters, we’ll shift our decimal place one, two, three spots to the left, giving
us a result of 0.310 meters.
And now, let’s look at the units in
the numerator and denominator of this fraction. We see, first of all, that these
units of meters in numerator cancel with meters here. And then also, the unit of teslas
cancels out from top and bottom, as does the unit of amperes because the ampere is
in the numerator and denominator, we could say, of our overall denominator. So just like we would hope for,
this result is going to be unitless because we’re calculating a pure number.
Now, when we enter this fraction on
our calculator, we find that we don’t actually get a whole number result. This may happen practically due to
the construction of a solenoid where, say, at the ends of the solenoid, one complete
turn may not be finished. So, there’s nothing necessarily
wrong that 𝑁 is not a whole number. But our question statement does
tell us to round our result to the nearest whole number. When we do this, we find a result
of 211. This is the number of turns this
solenoid has, to the nearest whole number.
Let’s summarize now what we’ve
learned about the magnetic field due to a current in a solenoid. In this lesson, we learned that a
solenoid is a wire that’s arranged in a series of loops or turns. We saw further that when a solenoid
carries an electric current, it creates a magnetic field. That field is essentially constant
within the loops of the solenoid while outside those loops, the field is much weaker
and of varying direction.
Within the loops of the solenoid,
the strength of the magnetic field 𝐵 is given by 𝜇 naught, the permeability of
free space, times the number of turns in the solenoid times the current magnitude
that it carries all divided by its total length. We saw further that this is equal
to 𝜇 naught times lowercase 𝑛 times 𝐼, where lowercase 𝑛 is the turn
density. This is a summary of the magnetic
field due to a current in a solenoid.
