Question

An airplane with a wingspan of 60.0 m flies parallel to the Earth's surface at a point where the downward component of the Earth's magnetic field is 0.40010⁻⁴T. If the induced potential between wingtips is 0.900 V, what is the plane's speed?

Answer 1

The plane's speed can be calculated using the formula for the Hall emf. Using the given values, the speed is determined to be 0.40010^-4 m/s.

Step-by-step explanation:

In this question, we are given the wingspan of an airplane, the downward component of the Earth's magnetic field, and the induced potential between the wingtips. We need to determine the plane's speed. To do this, we can use the formula for the Hall emf:

Hall emf = (BdV)/v

Where B is the magnitude of the magnetic field, d is the wingspan, V is the induced potential, and v is the speed of the plane. Rearranging the formula, we can solve for v:

v = (BdV)/Hall emf

Plugging in the given values, we have:

v = (0.40010^(-4) T)(60.0 m)(0.900 V)/(0.900 V)

Simplifying the equation, we find that the plane's speed is 0.40010^(-4) m/s. Therefore, the answer is 0.40010^(-4) m/s.