Video Transcript
A rectangular loop of conducting
wire moves at a constant speed perpendicularly to a 35-millitesla uniform magnetic
field that is 34 centimeters wide, as shown in the diagram. The loop has a resistance of 2.5
ohms. What is the average power of the
electric current induced in the loop as the loop moves through the field? Give your answer in scientific
notation to one decimal place.
The question states that current is
induced in a loop by the motion of the loop into and then out of a uniform magnetic
field. The question asked for the average
value of power, which is the rate at which energy is dissipated by the resistance of
the loop to the induced current. The average value of power is all
we can solve for, as the rate of dissipation of energy for the loop changes as the
loop moves and so does not have a constant value.
Clearing space on screen, Ohm’s law
states that the current in a circuit that has a potential difference across it is
given by the potential difference divided by the resistance. Our loop does not have a potential
difference across it exactly, but the motion of the loop in the magnetic field
exerts forces on free electrons in the loop, doing work on the free electrons. The number of joules of work done
per coulomb of charge of free electrons is equal to the induced emf in the loop due
to its motion in the magnetic field. We can say then that 𝜀, the
induced emf, equals 𝐼 times 𝑅.
Power is the rate at which work is
done. For the loop, this can be written
as 𝑃 equals 𝜀 squared divided by 𝑅. Recall also that mechanical power
is the rate at which work is done on an object by a force. Mechanical power equals the rate of
change of displacement of the object multiplied by the force acting on the object in
the direction of the displacement. In other words, power equals force
times velocity.
Clearing more space to work, let’s
imagine our wire in this position relative to the magnetic field. We’ll say that the right side of
the loop enters the magnetic field at the instant when time equals zero seconds. As the wire enters the magnetic
field, the magnetic flux density around the wire changes from being zero to being
nonzero. The change in the magnetic flux
density exerts forces on the free electrons in the wire, resulting in the free
electrons having a net motion toward one end of the wire.
If the right-hand wire was not
connected to other wires to form a loop, then the changes in the concentrations of
free electrons would not produce a current, only a potential difference across the
ends of the wire. The right-hand wire is part of a
loop however. Therefore, the greater
concentration of free electrons at one end of the wire repels free electrons in the
adjacent wire of the loop from that end. The lesser concentration of free
electrons at the opposite end of the wire attracts free electrons from the adjacent
wire in the loop toward that end. The result is a net flow of free
electrons around the loop, which is an electric current. The magnitude of the current
induced by the motion of the loop into the field can be determined.
Let’s also note that the magnetic
field, which we remember is uniform, has a width of 34 centimeters and is less than
the length of the loop, which is 86 centimeters. The current induced due to the
motion of the loop into the magnetic field only persists while the right-hand side
of the loop is inside the magnetic field and the left-hand side of the loop remains
outside the magnetic field. If both the right-hand and
left-hand sides of the loop are outside of the magnetic field, then the only sides
of the loop within the magnetic field are parallel to the motion of those sides. This means the potential
differences induced across these wires are not directed along the lengths of the
wires, and the potential differences induced have the same direction.
This combination of potential
differences does not result in a net flow of free electrons in the loop. Because the magnetic field is 34
centimeters wide, current is induced while the loop moves a distance of up to 34
centimeters into the field. No current is induced after this
until the right-hand side of the loop starts to move out of the magnetic field. The current induced due to the
motion of the loop out of the magnetic field only persists while the right-hand side
of the loop is outside the magnetic field and the left-hand side of the loop remains
inside the magnetic field.
The current induced has the same
magnitude as the current induced as the loop entered the magnetic field, but the
direction of the current is opposite to the direction of the current induced as the
loop entered the magnetic field. The current induced due to the
motion of the loop out of the magnetic field therefore continues until the loop has
moved another 34 centimeters.
Understanding this, we can now
begin to calculate the average power dissipated. First, let’s calculate the
magnitude of emf induced in the loop when either its right or its left side is
passing through the magnetic field. Since the magnetic field and the
velocity of the loop are perpendicular, we can apply the equation 𝜀 equals 𝐵 times
𝑣 times 𝑙 to determine the emf induced across a given side. In this equation, 𝐵 is the
magnetic field strength, 𝑣 is the loop’s speed, and 𝑙 is the length of the side of
the loop exposed to the magnetic field.
While the right side of the loop is
in the magnetic field, the loop’s speed is 12 centimeters per second, or 0.12 meters
per second, the length of wire in the field is 51 centimeters, or 0.51 meters, and
the magnetic field strength is given as a constant 35 milliteslas, or 0.035
teslas. This comes out to 0.002142
volts. Recall that this is the emf
magnitude when either the right or left side of the loop moves through the magnetic
field.
We can use this result to calculate
the power dissipated by recalling that power equals emf squared divided by
resistance. The resistance of the loop is given
as 2.5 ohms. So, the power dissipated is 1.835
and so on times 10 to the negative six watts. To find the average power
dissipated like our question asks, we need to consider the entire time interval for
which the loop moves.
Remembering that time equals
distance divided by speed, we can say that the right side of the loop is passing
through the magnetic field for a time, we’ll call it 𝑡 sub 𝑅, of 0.34 meters
divided by 0.12 meters per second, which equals 2.83 repeating seconds. During this time interval, the
power dissipated is what we found earlier approximately 1.835 times 10 to the
negative six watts.
Next, there’s an interval where
neither the right nor the left side of the loop is in the field. Clearing some space, let’s call
this time interval 𝑡 sub 𝑁, since it is when neither side of the loop is in the
field. 𝑡 sub 𝑁 equals 0.86 meters minus
0.34 meters all divided by 0.12 meters per seconds. This comes out to 4.3 repeating
seconds.
Lastly, we solve for the time
interval during which the left side of the loop is passing through the magnetic
field. This time, we’ll label it 𝑡 sub
𝐿, is the same as 𝑡 sub 𝑅 since the loop moves at a constant speed, 2.83
repeating seconds.
Clearing a bit more space to work,
we can now solve for the total time interval of interest, 𝑡 sub 𝑇. It’s equal to the sum of 𝑡 sub 𝑅,
𝑡 sub 𝑁, and 𝑡 sub 𝐿, exactly 10 seconds. The power dissipated is 1.835 times
10 to the negative six watts for a time of 𝑡 sub 𝑅 plus 𝑡 sub 𝐿, or 5.6
repeating seconds. Otherwise, for 4.3 repeating
seconds, the power dissipated is zero watts.
The average power dissipated is the
fraction of total time during which nonzero power was being dissipated multiplied by
that power 𝑃. Notice that in this expression the
units of seconds cancel out, leaving us with units of watts. Rounding this result to one decimal
place gives 1.0 times 10 to the negative six watts. This is the average power
dissipated in the loop as it moves across the magnetic field.
