Q7.53P - Introduction to Electrodynamics

Question

The current in a long solenoid is increasing linearly with time, so the flux is proportional $t$ :.$\varphi =\alpha t$Two voltmeters are connected to diametrically opposite points (A and B), together with resistors ( $R_1$and $R_2$ ), as shown in Fig. 7.55. What is the reading on each voltmeter? Assume that these are ideal voltmeters that draw negligible current (they have huge internal resistance), and that a voltmeter register -$-{\int }_a^{b}E\times dl$between the terminals and through the meter. [Answer: $V_1=\alpha R_1/(R_1+R_2)$. Notice that $V_1\ne V_2$ , even though they are connected to the same points]

Step-by-Step Solution

Answer

The expression for voltage across the resistance is $\frac{\alpha R_1}{R_1+R_2}$ and is $-\frac{\alpha R_2}{R_1+R_2}$.

Step 1: Write the given data from the question.

The relationship between current and flux,$\varphi =\alpha t$

The two registers are $R_1$ and $R_2$.

Step 2: Determine the formulas to calculate the voltmeter reading.

The expression for the induced emf is given as follows.

                    $\epsilon =-\frac{d\varphi }{dt}$                                            ……. (1)

Step 3: Draw the expression for the voltmeter reading.

Calculate the induced emf.

Substitute $\alpha t$ for $\varphi$ into equation (1).

$\begin{array}{l}\epsilon =\frac{d}{dt}\left(\alpha t\right)\\ \epsilon =\alpha \end{array}$

 

The current in the registers is given by

$I=\frac{\epsilon }{R_1+R_2}$

Substitute $\alpha$ for $\epsilon$ into above equation.

$I=\frac{\alpha }{R_1+R_2}$

 

The voltage across the register $R_1$ is given by,

$V_1=I{R}_1$

Substitute $\frac{\alpha }{R_1+R_2}$ for $I$ into above equation.

$\begin{array}{l}{V}_1=\frac{\alpha }{R_1+R_2}{R}_1\\ V_1=\frac{\alpha R_1}{R_1+R_2}\end{array}$

 

The voltage across the register $R_1$ is given by,

$V_2=-I{R}_2$

Substitute $\frac{\alpha }{R_1+R_2}$ for $I$ into above equation.

$\begin{array}{l}{V}_2=-\frac{\alpha }{R_1+R_2}{R}_2\\ V_2=-\frac{\alpha R_2}{R_1+R_2}\end{array}$

 

Hence, the expression for voltage across the resistance $R_1$ is $\frac{\alpha R_1}{R_1+R_2}$  and $R_2$ is $-\frac{\alpha R_2}{R_1+R_2}$.