Question

The circuit shown in the figure consists of two resistances $R_1$ & $R_2$ connected to two ideal voltmetres $V_1$ & $V_2$ . Assume that a voltmeter reads $\Delta V=-\underset{a}{\overset{b}{\int }}\vec{E}.d\vec{l}$ between its terminals. A time-varying magnetic field $B(t)=B_{0}t$ (where $B_0$ is a positive constant of proper dimensions and $t$ is time) exists in a circular region of radius a and it is directed into the plane of the figure. The reading of voltmeter $V_2$ is

• A. $\frac{\pi a^2{B}_0{R}_1}{R_1+R_2}$
• B. $-\frac{\pi a^2{B}_0{R}_2}{R_1+R_2}$
• C. $-\frac{\pi a^2{B}_0{R}_1}{R_1+R_2}$
• D. $\frac{\pi a^2{B}_0{R}_2}{R_1+R_2}$

Answer 1

Answer: (B)

$\varphi =\pi a^2{B}_{0}t$
$ϵ=-\frac{d\varphi }{dt}$
$\Rightarrow ϵ=\pi a^2{B}_0$ (induced emf)
$i=\frac{\pi a^2{B}_0}{R_1+R_2}$ (induced current)
$\Delta V_1=-\underset{a}{\overset{b}{\int }}\vec{E}\cdot d\vec{l}=-\left(-i{R}_1\right)=\frac{\pi a^2{B}_0{R}_1}{R_1+R_2}$
$\Delta V_2=-\underset{a}{\overset{b}{\int }}\vec{E}\cdot d\vec{l}=-\left(i{R}_2\right)=-\frac{\pi a^2{B}_0{R}_2}{R_1+R_2}$