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What is the emf produced in a 1.5 meter wire moving at a speed of 6.2 meters/second perpendicular to a magnetic field of strength 3.96 × 10-3newtons/amp·meter?
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What is the emf produced in a 1.5 meter wire moving at a speed of 6.2 meters/second perpendicular to a magnetic field of strength 3.96 × 10-3newtons/amp·meter?

User Jeremy Bourque
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2 Answers

0 votes

Answer:

For plato users: Option E.

Step-by-step explanation:

User Jurgispods
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3 votes
General formula for emf is : emf = vBL (sin 0)...(1)
As the angle here is 90º and sin90º =1.
So, equation (1) becomes : emf = vBL Putting values :
emf = (6.2)(1.5)(3.96 * 10^-3) = 0.0369 volts
Hope this helps:)

User Nathan Fisher
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