A small airplane with a wingspan of 13.5 m is flying due north at a speed of 70.8 m/s over a region where the vertical component of the Earth's magnetic field is 1.20 µT downward.
(a) What potential difference is developed between the airplane's wingtips?
mV
(b) Which wingtip is at higher potential?
The wingtip on the pilot's left.
The wingtip on the pilot's right.
(c) How would the answers to parts (a) and (b) change if the plane turned to fly due east?
Part (a) would increase and part (b) would reverse.
Part (a) would increase and part (b) would not change.
Part (a) would decrease and part (b) would reverse.
Part (a) would decrease and part (b) would not change.
Neither part would change.
(d) Can this emf be used to power a lightbulb in the passenger compartment?
Yes
No
Explain your answer.
To find the potential difference developed between the airplane's wingtips, we can use Faraday's law of electromagnetic induction. The formula to calculate the induced emf (potential difference) is given by:
emf = ΔFlux/Δt,
where ΔFlux represents the change in magnetic flux, and Δt represents the change in time.
In this case, we are given the wingspan of the airplane, the speed of the airplane, and the vertical component of Earth's magnetic field. Since the airplane is flying due north, we can assume that the wings are oriented east-west horizontally.
To calculate the change in magnetic flux, we need to determine the rate at which the magnetic field passing through the wings changes. Since the wingspan is fixed, we have a constant area, and the only change will occur due to the airplane's motion through the magnetic field.
ΔFlux = B * ΔA = B * v * Δt,
where B is the vertical component of Earth's magnetic field, v is the velocity of the plane, and Δt is the time interval in which the plane travels a distance equal to its wingspan.
Substituting the given values:
ΔFlux = (1.20 µT) * (70.8 m/s) * (13.5 m / 70.8 m/s) = 0.2448 T*m²/s.
Now, we can calculate the potential difference by dividing the change in flux by the change in time:
emf = ΔFlux/Δt = (0.2448 T*m²/s) / (Δt).
Since the time interval is not provided, we cannot find the exact value of the potential difference in millivolts (mV) without that information.
For part (b), the wingtip on the pilot's left will be at higher potential because the induced emf will cause an electric field to be established in the wing connecting the left and right wingtips. The electric field will point from higher to lower potential, so the left wingtip will be at higher potential.
In part (c), if the plane turned to fly due east, the magnitude of the emf (potential difference) would increase because the change in magnetic flux would be greater. However, the direction of the induced electric field would be reversed. Therefore, the answer to part (a) would increase, and the answer to part (b) would reverse because the left wingtip would then be at lower potential.
As for part (d), the induced emf generated between the airplane's wingtips can be considered as a potential energy source. However, whether it can power a lightbulb in the passenger compartment depends on the magnitude of the emf and the power requirements of the lightbulb. Without additional information, we cannot definitively say if it would be enough to power a lightbulb.