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 1

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.

Hope This Helps You!
Good Luck Studying ^-^

Answer 2

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

Explanation :

Induced EMF formula :

`e=Blv`

where,

e = emf

B = magnetic field =
`3.96* 10^(-3)Newtons/amp.meter`

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.

`e=(3.96* 10^(-3))* (1.5)* (6.2)=0.037V`

Therefore, the emf is, 0.037 volts