Video Transcript
A magnetic field of strength 10
times 10 to the negative five teslas is measured at a perpendicular distance of 12
centimeters from a long, straight wire. Some time later, a magnetic field
strength of 20 times 10 to the negative five teslas is measured at a perpendicular
distance of six centimeters from the same wire. Assuming no other changes to the
system, which of the following statements describes the magnitude of the current
being carried by the wire between the two measurements? (A) The current being carried by
the wire remains the same between the first and second measurements. (B) The current being carried by
the wire increases between the first and second measurements. Or (C) the current being carried by
the wire decreases between the first and second measurements.
In this question, we have a long,
straight current-carrying wire. We’re given information about the
magnetic field strength as measured at different distances from the wire at two
different times. We need to work out whether the
magnitude of the current carried by the wire changed between these two times. Let’s start by clearing some room
on screen and thinking about the first measurement of the magnetic field
strength.
We were told that at a point 12
centimeters from the wire, the strength of the magnetic field is measured as 10
times 10 to the negative five teslas. So, what does this tell us about
the magnitude of the current in the wire? Well, let’s recall that the
strength of a magnetic field at a point near a current-carrying wire is given by the
following equation. 𝐵 equals 𝜇 naught times 𝐼
divided by two 𝜋𝑑, where 𝐵 is the magnetic field strength, 𝐼 is the current in
the wire, 𝑑 is the distance between the wire and the point where the magnetic field
is measured, and 𝜇 naught is a constant called the permeability of free space.
In this question, we’re given the
values of both 𝐵 and 𝑑. And we want to use them to find
information about the current, 𝐼. So let’s rearrange the equation to
make 𝐼 the subject. We can do this by multiplying both
sides of the equation by two 𝜋𝑑 and dividing by 𝜇 naught. This leaves us with the equation 𝐼
equals two 𝜋𝑑 times 𝐵 divided by 𝜇 naught.
For this first measurement, we’re
told that the magnetic field strength, which we’ll call 𝐵 sub one, is equal to 10
times 10 to the negative five teslas at a distance of 12 centimeters from the
wire. We need to convert this distance
into plain meters before we can substitute into our equation, which gives us 𝑑 sub
one equals 0.12 meters. We also need to recall that the
permeability of free space, 𝜇 naught, has a value of four 𝜋 times 10 to the
negative seven tesla meters per ampere. Now, substituting these values into
the equation for 𝐼 and plugging the expression into a calculator, we reach a
current value of 60 amps. We’ll call this initial current 𝐼
sub one.
Let’s now work out the current in
the wire for the second measurement, when we know the magnetic field strength 𝐵 sub
two is measured as 20 times 10 to the negative five teslas at a distance of six
centimeters from the wire. Again, we need to convert this
distance into plain meters. So we write that 𝑑 two equals 0.06
meters. We’re ready to substitute these
values into our equation for 𝐼 and find the current in the wire at this later
moment in time. We’ll call this current 𝐼 two. And its value comes out to 60 amps,
the same as 𝐼 one. So, at both of the times when the
measurements are made, the wire carries a current of the same magnitude, 60
amps.
At first, this may seem
surprising. We have two different magnetic
field strengths, measured at two different distances. So how could these two sets of
values correspond to the same current? To understand this, let’s return to
our equation for the current: 𝐼 equals two 𝜋 𝑑 times 𝐵 divided by 𝜇 naught. Notice that the values two 𝜋 and
𝜇 naught are just constants. So, we can choose to simplify this
expression by ignoring them and rewriting our equation as a proportionality
relation: 𝐼 is proportional to 𝑑 times 𝐵.
For the first set of measurements,
which we’ve labeled with a subscript one, we have that 𝑑 one is equal to 0.12
meters and 𝐵 one is equal to 10 times 10 to the negative five teslas. We also know that 𝐼 one is
proportional to 𝑑 one times 𝐵 one. For the second set of measurements,
which we’ve labeled with a subscript two, we know that 𝐼 two is proportional to 𝑑
two times 𝐵 two, where 𝑑 two is 0.06 meters and 𝐵 two is 20 times 10 to the
negative five teslas.
If we compare these sets of
measurements, we might notice something interesting. 𝑑 two is equal to one-half of 𝑑
one, and 𝐵 two is equal to 𝐵 one multiplied by two. If we substitute these values into
our expression for 𝐼 two, we find that these factors of two cancel out. This means that both currents are
proportional to 𝑑 one times 𝐵 one. This is why we calculated the same
value for both currents.
Although 𝐵 two is greater than 𝐵
one, 𝑑 two is less than 𝑑 one. And the differences in both values
here cancel each other out. When the second measurement of the
magnetic field was made, the value of the field strength only increased because it
was measured closer to the wire. The magnitude of the current in the
wire did not change. These observations agree with
answer option (A). So this is our final answer. The current being carried by the
wire remains the same between the first and second measurements.
