Question

An astronaut who is repairing the outside of her spaceship accidentally pushes away a 92.9 cm long steel rod, which flies off at 12.5 m/s , never to be seen again. As it happens, the rod is oriented perpendicularly to the magnetic field in that region of space. The rod is moving perpendicularly to its length as well as to the direction of the magnetic field. The magnetic field strength there is 6.23 mT . What is the magnitude of the EMF, in millivolts, induced between the ends of the rod

Answer 1

Answer:

V = 0.0723 volts = 72.3 milivolts

Explanation:

The emf induced in the rod is the motional emf due to the magnetic field. This motional emf can be calculated by the following formula:

$EMF = V = vBl Sin\theta$

where,

V = Motional EMF = ?

v = speed of rod = 12.5 m/s

B = Magnetic Field = 6.23 mT = 0.00623 T

l = Length of rod = 92.9 cm = 0.929 m

θ = angle between v and B = 90°

Therefore,

$V = (12.5\ m/s)(0.00623\ T)(0.929\ m)Sin\ 90^o$

V = 0.0723 volts = 72.3 milivolts