A uniform magnetic field of \( 0.4 \) T acts perpendicular to a circular copper disc \( 20 \) cm in radius. The disc is having a uniform angular velocity of \( 10\pi \) rad/s about an axis through its center and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (\(\pi = 3.14\))
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For problems involving rotating conductors in a magnetic field, use \( V = \frac{1}{2} B\omega R^2 \), considering the radial motion of charge carriers.
The induced potential difference \( V \) in a rotating conducting disc is calculated using the formula \( V = \frac{1}{2} B \omega R^2 \). The given parameters are magnetic field strength \( B = 0.4 \) T, angular velocity \( \omega = 10\pi \) rad/s, and disc radius \( R = 0.2 \) m. Substituting these values into the formula yields \( V = \frac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2 = 0.08\pi \) V. With \( \pi \approx 3.14 \), the potential difference is \( V = 0.08 \times 3.14 = 0.2512 \) V. Therefore, the calculated induced potential difference is 0.2512 V.
