Question

A gramophone disc of brass of diameter 30 cm rotates
horizontally at the rate of 100/3 revolutions per minute. If
the vertical component of the earth's magnetic field be 0.01
weber/metre', then the emf induced between the centre
and the rim of the disc will be-
(a) 7.065 x 10^-4 v
(b) 3.9 * 10^-4 v
(c) None of the above
(d) 2.32 x 10^-4 V​

Answer 1

The emf induced between the center  and the rim of the disc will be $3.92\times 210^{-4}\ Volts$.

Explanation:

It is given that,

Diameter of the disc, d = 30 cm

Radius, r = 15 cm = 0.15 m

Angular velocity of the disc, $\omega=\dfrac{100}{3}\ rev/min$

The vertical component of the earth's magnetic field be 0.01 W/m. We need to find the emf induced between the center  and the rim of the disc. It is given by the product of magnetic flux and the number of revolution per second.

$\epsilon=\phi\times \dfrac{100}{60\times 3}$

$\epsilon=B\times \pi r^2\times \dfrac{100}{60\times 3}$

$\epsilon=0.01\times \pi (0.15)^2\times \dfrac{100}{60\times 3}$

$\epsilon=3.92\times 210^{-4}\ Volts$

So, the emf induced between the center  and the rim of the disc will be $3.92\times 210^{-4}\ Volts$. Hence, this is the required solution.

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