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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 ?

a.37 volts
b.0.0026 volts
c.5.0 volts
d.0.37 volts
e.0.037 volts

Respuesta :

Hello there.

Question: 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 ?

a.37 volts
b.0.0026 volts
c.5.0 volts
d.0.37 volts
e.0.037 volts

Answer: It would be E. 0.037.

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Answer : The correct option is, (e) 0.037 volts

Explanation :

Induced EMF formula :

[tex]e=Blv[/tex]

where,

e = emf

B = magnetic field = [tex]3.96\times 10^{-3}Newtons/amp.meter[/tex]

l = length of the conductor = 1.5 meter

v = speed of the conductor = 6.2 meter/second

Now put all the given values in the above formula, we get the emf.

[tex]e=(3.96\times 10^{-3})\times (1.5)\times (6.2)=0.037V[/tex]

Therefore, the emf is, 0.037 volts