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.
Fig. A uniform magnetic flux density perpendicular to the plane of the circle is changing at a rate [ 
= dB/dt]. What will be the reading of the voltmeter if the voltmeter is positioned: On the chord joining the two points of attachment?
Text Solution
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Sol. 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 RI. In both cases, applying Kirchhoff’s laws yields the equations:
RI + 
r(I + i) = 
a 2 
λ ,
ri + 
rI = π a 2 
.
In case , λ = θ and solution of the simultaneous equations shows that I, and hence the voltmeter reading, is zero.
In case , λ = θ – sin θ , after slightly lengthy algebra the voltmeter reading comes out to be
| V | = 
.
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