A 2.05 m length of wire is held in an east–west direction and moves horizontally to the north with a speed of 14.7 m/s. The vertical component of Earth's magnetic field in this region is 40.0 ?T directed downward. Calculate the induced emf between the ends of the wire and determine which end is positive.  magnitude     mV positive end

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A: Write a given values of this question.

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A: The induced emf is given by the rate of change of magnetic flux. e =dϕdt= BdAdt

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A: A = 0.160 m2 B = 0.403e-t/2 t = 4.5 s = ?

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A: The solution can be given as :-

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Q: A rectangular loop of area A=0.160 m2 is placed in a region where the magnetic field is…

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A: The length=2.00m  The magnitude =50.0mT

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A: The equation for the emf induced is

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