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 $∆V=-\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\overrightarrow{\mathit{E}}.\mathrm{d}\overrightarrow{l}$ between its terminals. A time-varying magnetic field $B\left(t\right)=B_{0}t$ (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

  $\frac{\mathrm{\pi }{a}^2{B}_0{R}_1}{R_1+R_2}$
• Correct answer
  B

  $-\frac{\mathrm{\pi }{a}^2{B}_0{R}_2}{R_1+R_2}$
• C

  $-\frac{\mathrm{\pi }{a}^2{B}_0{R}_1}{R_1+R_2}$
• D

  $\frac{\mathrm{\pi }{a}^2{B}_0{R}_2}{R_1+R_2}$

Solution

$\varphi =\mathrm{\pi }{a}^2{B}_{0}t$

$\mathrm{\epsilon }=-\frac{\mathrm{d}\varphi }{\mathrm{d}t}$

$\Rightarrow \mathrm{\epsilon }=\mathrm{\pi }{a}^2{B}_0$ (indeed emf)

$i=\frac{\mathrm{\pi }{a}^2{B}_0}{R_1+R_2}$ (induced current)




$∆V_1=-\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\overrightarrow{E}.\mathrm{d}\overrightarrow{\mathit{l}}=-\left(-i{R}_1\right)=\frac{\mathrm{\pi }{a}^2{B}_0{R}_1}{R_1+R_2}$

$∆V_2=-\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\overrightarrow{E}.\mathrm{d}\overrightarrow{\mathit{l}}=-\left(-i{R}_2\right)=-\frac{\mathrm{\pi }{a}^2{B}_0{R}_2}{R_1+R_2}$

Hence, the correct answer is option: 'B'.