I am confused by the following example in my textbook.
Question:-
A metallic rod of length $l$ is rotated with a angular velocity $\omega$, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $l$, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field $B$ is present everywhere and is parallel to the axis. What is the emf between the centre and the metallic ring ?
Solution :-
The area of sector traced by the rod in time $t$ is $\dfrac12 l^2 \theta$, where $\theta = \omega t$. Therefore $\varepsilon = \dfrac{B\omega l^2}{2}$.
The math in the solution is not hard to understand. I have trouble understanding the physics.
I know $\varepsilon = \dfrac{\mathrm d\ \phi_B}{\mathrm d \ t} = \dfrac{\mathrm d\ (\bf B\cdot A )}{\mathrm d \ t}$ and in the question it is given that ${\bf B}$ is constant and the angle between $\bf A$ and $\bf B$ is $0$. So we are just left with $\varepsilon = B\dfrac{\mathrm d \ A}{\mathrm d \ t}$.
Why the area taken in the solution is the area of the sector traced by the rod in time $t$ ?
Since the magnetic flux is passing through the whole metallic ring, shouldn't the area be the area of metallic ring ? Which is contant. So the emf should be zero.
Answer
To measure the EMF between the axial point and the metallic circumference we have to take a certain point on the circumference. Now, draw a line each from the axial point and the chosen point on the circumference. Join them and connect a galvanometer in parallel. Observe carefully here, w.r.t the point on the circumference, as the rod moves, the area under the circuit changes. So, according to EMI, there is an induced EMF. The change in area is relative. According to you the area was constant because you didn't consider any circuit.
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