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?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

Zero

None of these

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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.