In the August \(1992\) space shuttle flight, only \(250 m\) of the conducting tether considered in Example \(23.2\) could be let out. A \(40.0 V\) motional emf was generated in the Earth’s \(5.00 \times 10^{-5} \mathrm{~T}\) field, while moving at \(7.80 \times 10^{3} \mathrm{~m} / \mathrm{s}\) . What was the angle between the shuttle’s velocity and the Earth’s field, assuming the conductor was perpendicular to the field?

Equation Transcription:

Text Transcription:

1992

250 m

23.2

40.0 V

5.00 times 10^-5 T

7.80 times 10^3 m/s

Solution 22PE

=

= 24.2o

The angle between the velocity and the space shuttle is 24.2o