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 1 Wb $m^2$ 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. 230 V

Answer 1

Answer: (B)

0.23 V

Here,

$l=r=5cm=5\times {10}^{-2}m,$
B = 1 Wb $m^{-2}$

$\omega =2\pi \left(\frac{1800}{60}\right)rad{s}^{-1}=60\pi rad{s}^{-1},$

$ϵ=\frac{1}{2}B{l}^2\omega =\frac{1}{2}\times 1\times (5\times {10}^{-2}{)}^2\times 60\pi =0.23V$