Why Does Induced EMF Change in Generators but Not Moving Conductors?

[leojun]

why does induced emf change in case of generators but remains constant in case of moving conductor?

 

[leojun]

after thinking for a while,i myself came to the conclusion that constant change in magnetic flux(in moving conductor with constant vel) causes constant induced emf and sinusoidal change in magnetic flux(in rotating generator) causes sinusoidal induced emf...:);)

 

[Andrew Mason]

The second part of your question is a bit confusing. A conductor does not generate constant emf if it moves at constant speed through a constant magnetic field. There has to be a time rate of change of flux enclosed by the conductor.

Induced voltage is governed by Faraday's law: $\oint \vec{E}\cdot d\vec{s} = -\frac{d}{dt}\int \vec{B}\cdot d\vec{A}$

Since in a rotating generator, $\vec{B}\cdot d\vec{A}$ is integrated over the area enclosed by a conducting loop in the armature, the right side keeps changing if the loop rotates in a fixed magnetic field. So the induced voltage (the left side: the line integral of the electric field over the path around the loop) keeps changing.

If a conducting loop moves through a constant magnetic field there is no induced emf. This is because there is no change in the flux enclosed by the loop.

AM

 

[ehild]

If a conducting loop moves through a constant magnetic field there is no induced emf. This is because there is no change in the flux enclosed by the loop.

AM

Not the whole loop moves, but one side only in the experiments showing induced voltage in a moving straight piece of metal.

ehild

 

[leojun]

yup,the conductor here means the movable piece of metal...not the whole circuit.

 

[Andrew Mason]

Not the whole loop moves, but one side only in the experiments showing induced voltage in a moving straight piece of metal.

ehild

Ok. So the "moving conductor" is really an expanding loop in which dA/dt = velocity x length of the conductor. In that case $\xi = -\frac{d}{dt}\int\vec{B}\cdot d\vec{A}$ = -Bvl which is constant if v is constant.

AM

 

[ehild]

Ok. So the "moving conductor" is really an expanding loop in which dA/dt = velocity x length of the conductor. In that case $\xi = -\frac{d}{dt}\int\vec{B}\cdot d\vec{A}$ = -Bvl which is constant if v is constant.

AM

Yes.

ehild