Question

In a Faraday disc dynamo, a metal disc of radius \[R\] rotates with an angular velocity \[\omega \] about an axis perpendicular to the plane of the disc and passing through its center. The disc is placed in a magnetic field \[B\] acting perpendicular to the plane of disc. Determine the induced emf between the rim and the axis of the disc.

Answer 1

Hint: Use the formula for the magnetic flux and induced emf between the rim and axis of the disc. Convert the derived equation in terms of the angular velocity of the metal disc.

Formula used:
The magnetic flux is given by
\[\phi = BA\] …… (1)
Here, \[\phi \] is the magnetic flux, \[B\] is the magnetic field and \[A\] is the area.
The area of a circle is given by
\[A = \pi {R^2}\] …… (2)
Here, \[A\] is the area of the circle and \[R\] is the radius of the circle.
The induced emf is given by
\[e = \dfrac{{d\phi }}{{dt}}\] …… (3)
Here, \[e\] is the induced emf, \[d\phi \] is the change in the magnetic flux and \[dt\] is the change in the time.
The time period \[T\] of an object is given by
\[T = \dfrac{{2\pi }}{\omega }\] …… (4)
Here, \[\omega \] is the angular velocity of the object.

Complete step by step answer:
The metal disc of radius \[R\] rotates with an angular velocity \[\omega \] about an axis perpendicular to the plane of the disc and passing through its center.
The disc is placed in a magnetic field \[B\] acting perpendicular to the plane of disc.
Determine the change in flux.
\[d\phi = BdA\]
Here, \[dA\] is the area of the metal disc swept in one rotation.
Substitute \[\pi {R^2}\] for \[dA\] in the above equation.
\[d\phi = B\left( {\pi {R^2}} \right)\]
\[ \Rightarrow d\phi = \pi B{R^2}\]
Now determine the induced emf between the rim and the axis of the disc.
Substitute \[\pi B{R^2}\] for \[d\phi \] and \[T\] for \[dt\] in equation (3).
\[e = \dfrac{{\pi B{R^2}}}{T}\]
Here, \[T\] is the time period of the metal disc for one rotation.
Substitute \[\dfrac{{2\pi }}{\omega }\] for \[T\] in the above equation.
\[e = \dfrac{{\pi B{R^2}}}{{\dfrac{{2\pi }}{\omega }}}\]
\[ \Rightarrow e = \dfrac{1}{2}B{R^2}\omega \]
Hence, the induced emf between the rim and the axis of the disc is \[\dfrac{1}{2}B{R^2}\omega \].

Note:
The emf induced in the metal disc is during one rotation of the disc about its axis.
Due to the rotation of the metal disc in the magnetic field, there is production of changing flux which results in the induced emf.