To check this out, calculate the induced voltage for a wild goose with a wingspan of 1.5 m at level flight due south at 13 m/s at a point where the earth's magnetic field is 3 ×10−5 T directed downward from horizontal by 14 degrees. What would be the expected voltage difference across the goose's wingtips?

Complete question:

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been suggested that perhaps they detect a motional EMF as they fly north to south, but it turns out that the induced voltages are small compared to the voltages normally encountered in cells, so this is probably not the mechanism involved.

To check this out, calculate the induced voltage for a wild goose with a wingspan of 1.5 m at level flight due south at 13 m/s at a point where the earth's magnetic field is 3 ×10−5 T directed downward from horizontal by 14 degrees. What would be the expected voltage difference across the goose's wingtips?

The expected voltage difference across the goose's wingtips is 0.1415 mV

Explanation:

Given;

wingspan of the wild goose, L = 1.5 m

speed of the wild goose, v = 13 m/s

Earth's magnetic field strength, B =  3 × 10⁻⁵ T

angle of inclination of the bird's motion to Earth's magnetic field, θ = 14⁰

If the bird's wings act like conduct in the Earth's magnetic field, then induced emf according to Faraday, will be calculated as follows;

emf = BLvsinθ

where;

B is Earth's magnetic field strength

L is length of the conductor (wingspan)

v is velocity of the conductor (wingspan)

θ is the angle of inclination to the magnetic field

Induced emf = (3 × 10⁻⁵ )( 1.5 )(13 )sin14⁰

Induced emf = 1.4152 x 10⁻⁴ V = 0.1415 mV

Thus, the expected voltage difference across the goose's wingtips is 0.1415 mV

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.