The diagram shows a permanent
magnet being moved through a loop of copper wire. This motion induces an electric
current of 0.5 amperes in the wire. If the magnet is moved through the
loop at half the speed, what will the current in the loop be? If the permanent magnet is changed
for one that is twice as strong and moves through the loop at the original speed,
what will the current in the loop be?
Alright, looking over our diagram,
we see our loop of copper wire, a conducting material, and the permanent magnet
which is moving through it. We’re told that when this happens,
when the wire moves through the loop, it induces a current of 0.5 amperes in the
wire. This motion of the magnet moving
through the loop occurs at some speed. We can call it 𝑆, even though it’s
not labelled in our diagram.
Our first question asks if we
change nothing about our setup, except the speed with which we move the magnet; we
make it half as big as it was before, then what will happen to the current induced
in the loop? To start figuring this out, it will
be helpful to draw in the magnetic field lines that show the magnetic field created
by this magnet. That field and the field lines
representing it look something like this. So initially, we move this magnet
and its magnetic field through the loop at the speed we’re calling 𝑆.
That means that the total magnetic
field through this loop is changing while the magnet moves. And the rate of that change — the
speed with which it occurs—has to do with the speed 𝑆. The higher 𝑆 is, the faster the
magnet is moving and therefore the faster the magnetic field through the loop is
changing. And this change is the mechanism
that induces current in the wire. The rate at which that change
occurs directly corresponds with the amount of current induced. In other words, the faster the
magnetic field through the loop is changing, the more current will be induced in the
That bit is important because we’re
told that the modification in this first question is that we no longer move the
magnet with our speed 𝑆. The original speed will be moving
at half that speed. Since we’re moving the magnet
relatively more slowly, that means the magnetic field experienced by the loop will
change more slowly. When that rate of change of
magnetic field through the loop goes down, so will be induced current.
We don’t know exactly what the
current will be when we move our magnet at half the original speed. But we just know that it will be
less than the original amount of 0.5 amperes. We’ll write that down as our
answer. And the explanation for it like we
saw is that the rate of change of the magnetic field through our conducting loop is
decreasing relative to what it was originally. Less change means less induced
current, which means that whatever the current is will be less than 0.5 amperes.
In part two of our question, we
asked if the permanent magnet is changed out for one that’s twice as strong, but
moves through the loop at the original speed — what we’re calling 𝑆 — what will the
current in the loop be. If we were to double the strength
of this magnet and therefore the strength of its magnetic field, while keeping the
motion of the magnet the same, then the question is what effect will that have on
the rate of change of the magnetic field through this loop.
With the field of the magnet being
stronger overall, that means we could expect the change in magnetic field strength
from moving from one pole of the magnet to the other to be greater than it was
before. That means if we pass this magnet
entirely through the loop, the change in magnetic field experienced by the loop
would increase. That increase will lead to an
increase in induced electric current.
Just like before, we can’t say
exactly what the current will be in this modified case. But we do expect that it will be
greater than what it was before, greater than 0.5 amperes. And we write that down as our
answer because we’ve seen that the rate at which the magnetic field through the loop
changes is increased in this case. And we expect that increase to
increase the current induced.