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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 × 10-3 newtons/amp·meter ? 37 volts 0.0026 volts 5.0 volts 0.37 volts 0.037 volts?

User SubjectX
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2 Answers

3 votes
E. 0.037 Volts. It's correct for Plato. The actual answer is around 0.0369 Volts
User It All Makes Cents
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4 votes

Answer:

0.037 volts

Step-by-step explanation:

The formula for calculating the Electromagnetic field induction in a moving wire is:


Emf=B*l*v*sin(\theta )

Where:

B is the strength of the magnetic field in Webers or
(N*m)/(A)

l is the longitude of the wire in meters

v is the speed of the wire crossing the magnetic field in meters per second (m/s)

and
\theta is the angle between the magnetic field and the wire

So, substituting the values into the formula:


Emf = 3.96 * 10^(-3) W * 1.5 m * 6.2 (m)/(s) * sin(90)

sin(90)=1 so:


Emf = 3.96 * 10^(-3) W * 1.5 m * 6.2 (m)/(s) = 0.0368 volts ≅ 0.037 volts

User Alessandro Verona
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