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An astronaut, while repairing the outside of her spaceship, accidentally pushes away a 99.3-cm-long steel rod, which flies off at 13.3 m/s, never to be seen again. The rod is oriented perpendicularly to the magnetic field in that region of space and is moving perpendicularly to its length as well as to the direction of the magnetic field. The magnetic field strength is 6.23 mT.

What is the magnitude of the EMF, in millivolts, induced between the ends of the rod?

Answer :

Answer:

82.3mV

Explanation:

Detailed explanation and calculation is shown in the image below

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Rewritten by : Barada

Final answer:

The magnitude of the EMF induced between the ends of the steel rod is 8.29 millivolts.

Explanation:

The magnitude of the EMF induced between the ends of the steel rod can be calculated using Faraday's law of induction. The formula for the magnitude of the induced EMF is given by emf = Blv, where B is the magnetic field strength, l is the length of the rod, and v is the velocity of the rod. Plugging in the given values, we have emf = (6.23 mT)(99.3 cm)(13.3 m/s) = 8.29 mV.

So, the magnitude of the EMF induced between the ends of the rod is 8.29 millivolts.

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