Question

A circular coil is made from wire of uniform area of cross section and having resistance 12Ω. A voltmeter of resistance 10Ω is connected between the points $P$ and $Q$ of the coil and placed at the centre of the coil as shown in the figure. Given magnetic field within region, $B=2t$ tesla. Area of circular coil is 6 m² (Assume connecting wire has no resistance). Calculate the reading of the volt meter in volt.

Answer 1

2 V

Answer 2

We first note that the uniform magnetic field gives a flux

$$\Phi = B\cdot A = (2t)(6)=12t,$$

so the induced emf (in magnitude) in the loop is

$$\varepsilon = \frac{d\Phi}{dt} = 12\;{\rm V}\,.$$

Since the coil is made of wire of uniform resistance, the emf is “distributed” along it. The circular coil (360°) has a total resistance of 12 Ω so that the resistance of any arc is proportional to its subtended angle. In particular, the shorter arc $PQ$ (60°) has resistance

$$R_1 = \frac{60}{360}\times 12 =2\; \Omega\,,$$

and the rest of the coil (300°) has

$$R_2 = \frac{300}{360}\times 12 = 10\; \Omega\,.$$

In non‐conservative fields (as in induction) the “voltage” between two points becomes path–dependent. However if one uses a voltmeter whose leads are arranged so that they do not cut any changing flux (for example, by stretching them through the centre, where the field is essentially not “seen” by the connecting wires), the voltmeter “reads” the difference in the scalar potential which may be identified with the drop along one of the arcs. In other words, the “measured” potential difference between $P$ and $Q$ is given by the fraction of the total induced emf that falls along the arc between them. Choosing the arc of 60° (whose resistance is $2\;\Omega$) we have

$$V_{PQ}=\frac{R_1}{R_{\rm total}}\,\varepsilon = \frac{2}{12}\times 12 = 2~\text{V}\,.$$

Thus the voltmeter reads 2 volts.

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Minimal Explanation of the Core Solution

1. Flux through loop: $\Phi=12t$ so $\varepsilon=\dfrac{d\Phi}{dt}=12$ V.
2. Resistances: 60° arc: $R_1=2\;\Omega$; 300° arc: $R_2=10\;\Omega$.
3. With the leads taken via a region with no induced emf, the potential difference is the drop on the 60° arc: $(2/12)\times12=2$ V.