Question

A wedding ring is tossed into the air and given a spin, resulting in an angular velocity of 13.5 rev/s. The rotation axis is a diameter of the ring. The magnitude of the Earth's magnetic field is \(4.97 \cdot 10^{-5} \mathrm{~T}\) at the ring's location. If the maximum induced voltage in the ring is \(1.446 \cdot 10^{-6} \mathrm{~V},\) what is the diameter of the ring?

Short answer

Answer: The diameter of the spinning wedding ring is approximately \(0.07958 \ \text{m}\).

Step-by-step solution

Convert Angular Velocity

Given angular velocity \(w=13.5 \ \text{rev/s}\), to convert it into radians per second, we multiply it by \(2\pi\): \(w_{radian} = 13.5 \cdot 2\pi \ \text{rad/s}\)

Formula for Induced Voltage

We know the formula for induced voltage in a spinning circular loop is: \(V = \pi \cdot D^{2} \cdot B \cdot w / 4\)

Solve for Diameter

We need to solve for \(D\). Plug in the values we have for the induced voltage \((V = 1.446\cdot 10^{-6}\ \text{V})\), Earth's magnetic field \((B = 4.97\cdot10^{-5}\ \text{T})\), and angular velocity in radians per second: \(1.446 \cdot 10^{-6} = \pi \cdot D^{2} \cdot (4.97\cdot10^{-5}) \cdot(13.5\cdot 2\pi)/4\) Now divide both sides by \(\pi \cdot (4.97\cdot10^{-5}) \cdot (13.5\cdot 2\pi)/4\): \(D^{2} = \frac{1.446 \cdot 10^{-6}}{ \pi \cdot (4.97\cdot10^{-5}) \cdot (13.5\cdot 2\pi)/4}\) Calculate \(D^{2}\): \(D^{2} \approx 0.006333\) Take the square root of both sides to find the diameter of the ring: \(D = \sqrt{0.006333} \approx 0.07958\ \text{m}\) So, the diameter of the ring is approximately \(0.07958 \ \text{m}\).