# In a Faraday disc dynamo, a metal disc of radius R rotates with an angular velocity ω about an axis perpendicular

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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 centre. 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.

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Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction $$\vec{B}$$ (see the figure in the above Note for reference). $$\vec{B}$$ points downwards. Let the constant angular speed of the disc be &omega;.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2&pi;rdr. Therefore, the time rate at which the element traces out the area is
$$\frac{d A}{d t}$$ = frequency of rotation x dA = fdA
where f = $$\frac{\omega}{2 \pi}$$ is the frequency of rotation.
.&rsquo;. $$\frac{d A}{d t}=\frac{\omega}{2 \pi}$$ (2 &pi;r dr) = &omega;r dr
The total emf induced between the axle and the rim of the rotating disc is
$$|e|=\int B \frac{d A}{d t}=\int_{0}^{R} B \omega r d r=B \omega \int_{0}^{R} r d r=B \omega \frac{R^{2}}{2}$$
For anticlockwise rotation in $$\vec{B}$$ pointing down, the axle is at a higher potential.