Q49 / 50 · Manipal (MET) 2023 · Physics
A vertical disc of diameter 10 cm makes 20 revolutions per second about a horizontal axis passing through its centre. A uniform magnetic field $10^{-1} \mathrm{~T}$ acts perpendicular to the plane of the disc. If the potential difference between its centre and rim is found to be $\frac{\pi}{x} \times 10^{-p}$ Volt, then find the ratio of $x / p$.
Correct answer: 1
Solution
As we know, disc is equivalent to a rod of length $r$. Divide disc into rods, each rod can be replaced by a battery and these batteries are would be paralial in arrangement.
According to the question, radius of disc, $r=\frac{10}{2}=5 \mathrm{~cm}=0.05 \mathrm{~m}$ Angular velocity, $\omega=20 \times 2 \pi=40 \pi \mathrm{rad} / \mathrm{s}$ Magnetic field, $B=10^{-1} \mathrm{~T}$ The emf of a disc is given as $e=\frac{1}{2} B \omega r^2=\frac{1}{2}\left(10^{-1}\right)(40 \pi)(0.05)^2$ $\Rightarrow \quad V_0-V_A=1.57 \times 10^{-2} \mathrm{~V}$ $=\frac{\pi}{2} \times 10^{-2} \mathrm{~V}=\frac{\pi}{x} \times 10^{-p} \mathrm{~V}$ (given) $\therefore \quad \frac{x}{p}=\frac{2}{2}=1$
According to the question, radius of disc, $r=\frac{10}{2}=5 \mathrm{~cm}=0.05 \mathrm{~m}$ Angular velocity, $\omega=20 \times 2 \pi=40 \pi \mathrm{rad} / \mathrm{s}$ Magnetic field, $B=10^{-1} \mathrm{~T}$ The emf of a disc is given as $e=\frac{1}{2} B \omega r^2=\frac{1}{2}\left(10^{-1}\right)(40 \pi)(0.05)^2$ $\Rightarrow \quad V_0-V_A=1.57 \times 10^{-2} \mathrm{~V}$ $=\frac{\pi}{2} \times 10^{-2} \mathrm{~V}=\frac{\pi}{x} \times 10^{-p} \mathrm{~V}$ (given) $\therefore \quad \frac{x}{p}=\frac{2}{2}=1$
Hence, the correct answer is 1.