Calculate the motional emf induced along a 20.0 km long conductor moving at an orbital speed of 7.80 km/s perpendicular to the Earth’s 5.00 \times 10^{-5} T magnetic field.
Strategy
This is a straightforward application of the expression for motional emf— emf = Bℓv .
Entering the given values into emf = Bℓv gives
\operatorname{emf}=B t v (23.10)
=\left(5.00 \times 10^{-5} T \right)\left(2.0 \times 10^{4} m \right)\left(7.80 \times 10^{3} m / s \right)=7.80 \times 10^{3} V.
Discussion
The value obtained is greater than the 5 kV measured voltage for the shuttle experiment, since the actual orbital motion of the tether is not perpendicular to the Earth’s field. The 7.80 kV value is the maximum emf obtained when θ = 90º and sin θ = 1 .