Solution 3: Voltmeter Readings in Non-Conservative Field
1. Circuit Analysis:
A constant EMF of $\varepsilon = 12\,\text{V}$ is induced in the ring. The ring acts as a closed circuit with uniform resistance. The current circulating in the ring is $I$.
The points of connection separate the ring into two segments:
- Quarter arc (Minor arc): Length $L/4$, Resistance $R/4$.
- Three-quarter arc (Major arc): Length $3L/4$, Resistance $3R/4$.
The potential difference measured by a voltmeter in a time-varying magnetic field depends on the path of the voltmeter leads relative to the magnetic flux.
2. Calculation using Faraday’s Law on Loops:
We apply Kirchhoff’s law in the form $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi}{dt}$ for the loop formed by the voltmeter and the ring segment.
Case I (Figure I):
The voltmeter is connected across the minor arc (quarter circle). The wires run outside the magnetic field region without enclosing the central flux.
- The loop (Voltmeter + Minor Arc) encloses $\Phi = 0$.
- Equation: $V_{\text{reading}} – I(R_{\text{minor}}) = 0$ (Assuming ideal voltmeter reading balances potential drop).
- Technically: $\Delta V_{\text{ext}} = \Delta V_{\text{arc}}$. For a non-enclosing path, the reading is simply $I \times R_{\text{arc}}$.
- Current $I = \frac{\varepsilon}{R} = \frac{12}{R}$.
- Resistance of minor arc $R_{\text{minor}} = \frac{R}{4}$.
- $V_I = I \cdot R_{\text{minor}} = \left(\frac{12}{R}\right) \left(\frac{R}{4}\right) = 3\,\text{V}$.
Case II (Figure II):
The voltmeter is connected to the same points, but the wires loop around to enclose the magnetic field region (or connect across the major arc such that the loop composed of Voltmeter + Minor Arc encloses the flux).
Alternatively, consider the loop formed by the Voltmeter and the Major Arc (3/4 of ring). If the voltmeter is on the right, the loop with the Major Arc encloses no flux (assuming wires don’t cross the center).
- Resistance of Major Arc $R_{\text{major}} = \frac{3R}{4}$.
- $V_{II} = I \cdot R_{\text{major}} = \left(\frac{12}{R}\right) \left(\frac{3R}{4}\right) = 9\,\text{V}$.
Note: In induced electric fields, $V_{AB}$ is path-dependent. The voltmeter reading corresponds to the $IR$ drop of the path that completes a loop with the voltmeter without enclosing the changing magnetic flux.