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Question
In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity ω 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 the disc. Determine the induced emf between the rim and the axis of the disc.
Solution
Assume a thin conducting disc with radius R is rotated anticlockwise around its axis in a plane perpendicular to a uniform magnetic field of induction `vec "B"` (see the figure in the above Note for reference). The arrow `vec"B"` points downward. Let the constant angular speed of the disc be ω.
Consider an infinitesimal element with radial thickness dr at r from the rotation axis. The area traced by the element in one rotation is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is
`"dA"/"dt"` = frequency of rotation x dA = fdA
where f = `omega/(2pi)` is the frequency of rotation.
∴ `"dA"/"dt" = omega/(2pi) (2pi"rdr") = omega"rdr"`
The total emf induced between the axle and the rim of the rotating disc is
`|"e"| = int "B" "dA"/"dt"`
`= int_0^"R" "B"omega"rdr"`
`= "B"omega int_0^"R" "rdr"`
`= "B"omega "R"^2/2`
`= 1/2 ("B" omega"R"^2)`
For anticlockwise rotation in `vec"B"` pointing down, the axle is at a higher potential.
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