Question

A circular copper disc 10 cm in diameter rotates at 1800 revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction B of $1Wb/{{m}^{2}}$ is perpendicular to disc. What potential difference is developed between the axis of the disc and the rim?
A: 0.023 V
B: 0.23 V
C: 23 V
D: 230V

Answer 1

This situation is analogous to the case when a rod of a particular length and an angular velocity is rotated in a perpendicular field of a given induction. In such a scenario, an emf is generated and a similar value of emf is generated in this question. We can proceed to solve this question using that formula.

Formula used:
$E=\dfrac{1}{2}B{{l}^{2}}\omega$, where W is the EMF generated, B is the induction of the perpendicular magnetic field, l is the length of the rod and $\omega$is the angular velocity of the rod.

Complete step by step answer:
In the question we are given a circular copper disc that is 10 cm in diameter rotating at 1800 revolutions per minute about an axis through its centre and at right angles to the disc.
We are also given a uniform field of induction B of $1Wb/{{m}^{2}}$ which is perpendicular to disc.

We can find the angular velocity from the given frequency using the formula $\omega =2\pi \upsilon =2\pi (\dfrac{1800}{60})=60\pi rad/s$
The radius of the disc is taken as the length hence, $l=r=0.05m$
$B=1Wb/{{m}^{2}}$
$E=\dfrac{1}{2}B{{l}^{2}}\omega =\dfrac{1}{2}\times 1\times {{(0.05)}^{2}}\times 60\pi =0.23V$

So, the correct answer is “Option B”.

Note:
The above formula is derived from the concept where a rod that is of infinitesimally small length is taken. If it is at a distance of x from a circular path, we can find the emf generated along this element and upon integrating it to a whole, we get the value of the overall emf generated in the system.