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 potential difference \( V \) induced in a rotating conducting disc is determined by the formula:
\[
V = \frac{1}{2} B \omega R^2
\]
The given parameters are:
- \( B = 0.4 \) T (magnetic field strength),
- \( \omega = 10\pi \) rad/s (angular velocity),
- \( R = 20 \) cm \( = 0.2 \) m (disc radius).
Substituting these values into the equation:
\[
V = \frac{1}{2} \times 0.4 \times 10\pi \times (0.2)^2
\]
\[
V = \frac{1}{2} \times 0.4 \times 10\pi \times 0.04
\]
\[
V = \frac{1}{2} \times 0.4 \times 0.4\pi
\]
\[
V = \frac{1}{2} \times 0.16\pi
\]
\[
V = 0.08\pi
\]
Using the approximation \( \pi = 3.14 \), the calculation yields:
\[
V = 0.08 \times 3.14 = 0.2512 \, \text{V}
\]
Therefore, the calculated potential difference is 0.2512 V.
