### Related Formula The induced electromotive force (EMF) developed between the center and the rim of a rotating disc in a perpendicular magnetic field is given by: E = frac12 B omega R^2$E = \frac{1}{2} B \omega R^2$ ### Core Logic Given parameters from the problem statement [cite: 655, 657, 658]: * Magnetic field, B = 0.4 text T$B = 0.4 \text{ T}$ * Radius of the disc, R = 20 text cm = 0.2 text m$R = 20 \text{ cm} = 0.2 \text{ m}$ * Angular velocity, \omega = 10\pi \text{ rad s}^{-1} Substituting the values into the governing formula: E = frac12 times 0.4 times (10 times 3.14) times (0.2)^2$E = \frac{1}{2} \times 0.4 \times (10 \times 3.14) \times (0.2)^2$ E = 0.2 times 31.4 times 0.04$E = 0.2 \times 31.4 \times 0.04$ E = 0.2512 text V$E = 0.2512 \text{ V}$ ### Step 1: Evaluation The potential difference developed between the axis of the disc and the rim is precisely 0.2512 text V$0.2512 \text{ V}$. ### Pattern Recognition For any rotating conductor of length R$R$ or a continuous disc rotating about its center in a perpendicular magnetic field, the induced EMF is mathematically equivalent to a single radial rod sweeping the area, leading directly to the formula frac12Bomega R^2$\frac{1}{2}B\omega R^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction