A small airplane with a wingspan of 13.0 m is flying due north at a speed of 65.
ID: 1345704 • Letter: A
Question
A small airplane with a wingspan of 13.0 m is flying due north at a speed of 65.2 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 please.
Explanation / Answer
The B-field has a component parallel to the veocity of the wings, "B(y)= BCos(90)" and a component perpendicular to the velocity, "B(z) = BSin(90)" .
The B(y) component has no effect
The B(z) component exerts a force, F(x) on the charges in the wing which causes them to separate and migrate to the tips.
The force is;
F(x) = qvB(z)
The work done by this force in moving the charges a distance L is
W = F(x)L = qvLB(z)
The work is also the aquired potential energy change of the charges. The work per unit charge is the Potential Difference;
Potential Diff. = W/q = vLB(z) = vLBSin(90)
= (65.2)(13)(1.2x10^-6)Sin(90)
= 0.1*10-3 volts