I realise that similar questions have been asked on this site however these questions yielded answers which were not helpful . According to faradays law the induced emf in a conductor is directly proportional to the rate of change of magnetic flux .In the image below we see a direct motor armature , but let's just think of it to be an ac generator since it is the rotation of the armature that concerns me and not the configuration of the device .enter image description here

Moving from right to left . We see that in the first picture the coil is perpendicular to the flux , indicating maximum flux , let's take this to be our initial position hence the angle on our graph would be zero degrees . The induced emf is zero since we have not changed the magnetic flux with respect to time . As the armature moves to the second potion the flux has changed from maximum to minimum , indicating a large change in flux hence a large induced emf .Now as the coil moves from being parallel to the field ,back to being perpendicular again , the flux should change from being a minimum to a maximum , indicating a large induced emf , however the sine graph of induced emf against angle for an alternating current generator moves back down to being zero why is this so ? Please note I have no knowledge on angular momentum or torque .


The magnitude of the flux $\int \vec B\cdot d\vec A$ is maximum in figure A and C, but 0 in figure B and D because $\vec B$ is parallel to vector normal to the surface in $A$ and $C$ but perpendicular to the normal to the surface in situations B and D.

The flux $\Phi(t)$ is $$ \Phi(t)=\int_A \vec B\cdot d\vec A= A B \cos\omega t $$ since $\vert \vec B\vert$ is constant and the angle between $d\vec A$ and $\vec B$ changes as $\cos\omega t$.

Since the EMF ${\cal E}$ is $-d\Phi(t)/dt$ one immediately gets $$ {\cal E}=A B \omega \sin\omega t\, . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.