A uniform thin wire of length $2 π a$ and resistance $r$ has its ends joined to form a circle. A small voltmeter of resistance $R$ is connected by tight leads of negligible resistance to two points on the circumference of the circle at angular separation $θ$ , as shown in [Fig. (a) and (b)].
A uniform magnetic flux density perpendicular to the plane of the circle is changing at a rate $[\dot{B} = d B / d t]$ . What will be the reading of the voltmeter if the voltmeter is positioned at the center of the circle?

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$+ \frac{2 π R a^{2} \dot{B} \sin θ}{4 π^{2} R + r θ (2 π - θ)}$

Zero

$- \frac{2 π R a^{2} \dot{B} \sin θ}{4 π^{2} R + r θ (2 π - θ)}$

None of these

answer is B.

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Detailed Solution

Let the current through major arc be $I$ and that through voltmeter be $I$ . Using KVL according to sign convention to two different closed loops we can solve $I$ and $I$ . The required voltmeter reading is given by $R I$ . In both cases, applying Kirchhoff’s laws yields the equations:
$R I + \frac{θ}{2 π} r (I + i) = \frac{1}{2} a^{2} \dot{B} λ,$
$r i + \frac{θ}{2 π} r I = π a^{2} \dot{B} .$
In this case, $λ = θ$ and solution of the simultaneous equations shows that $I$ , and hence the voltmeter reading is zero.