Step 1: Understanding the Concept:
A rotating conducting disc in a perpendicular magnetic field behaves like an infinite collection of radial rods. Each rod induces an EMF, and since they are all in parallel, the total EMF between the center and the rim is that of a single radial rod.
Step 2: Key Formula or Approach:
Induced EMF \(e = \frac{1}{2} B \omega r^{2}\).
Where \(\omega = 2\pi f\) is the angular frequency and \(r\) is the radius.
Step 3: Detailed Explanation:
Given:
Diameter \(= 10\text{ cm} \implies r = 5\text{ cm} = 0.05\text{ m}\).
Frequency \(f = 20\text{ rev/s}\).
\(\omega = 2 \pi \times 20 = 40 \pi\text{ rad/s}\).
\(B = 10^{-1}\text{ T}\).
Substitute these into the EMF formula:
\[ e = \frac{1}{2} \times 10^{-1} \times 40 \pi \times (0.05)^{2} \]
\[ e = 2 \pi \times 10^{-1} \times 0.0025 = 2 \pi \times 10^{-1} \times (2.5 \times 10^{-3}) \]
\[ e = 5 \pi \times 10^{-4}\text{ V} \]
Rewriting in the form \(\pi/x \times 10^{-p}\):
\[ e = \frac{\pi}{2} \times 10 \times 10^{-4} = \frac{\pi}{2} \times 10^{-3} \]
Wait, let's re-verify: \(0.5 \times 0.1 \times 40\pi \times 0.0025 = 2\pi \times 0.0025 = 0.005\pi\).
\(0.005\pi = 5\pi \times 10^{-3} = \frac{\pi}{2} \times 10 \times 10^{-3} = \frac{\pi}{2} \times 10^{-2}\).
Comparing with \(\pi/x \times 10^{-p}\):
\(x = 2, p = 2\).
The ratio \(x/p = 2/2 = 1\).
Step 4: Final Answer:
The ratio \(x/p\) is 1.