Chapter 20, Problem 96P

(a)

To determine

When the loop of wire is rotated by $180°$, how much charge flows through the circuit?

Answer to Problem 96P

The charge that flows through the circuit is $8.5\text{C}$.

Explanation of Solution

Write the equation to find the induced emf in the loop.

$\epsilon =-N\frac{\Delta {\varphi }_B}{\Delta t}$ (I)

Here, $\epsilon$ is the induced emf, $N$ is the number of loops, $\Delta {\varphi }_B$ is the change in magnetic flux, $\Delta t$ is the change in time

The initial magnetic flux through the loop is $BA$ and the final magnetic flux through the loop is $-BA$.

Rewrite equation (I).

$\begin{array}{c}\epsilon =-N\frac{-BA-AB}{\Delta t}\\ =-\frac{2NBA}{\Delta t}\end{array}$ (II)

The emf in a circuit is the product of current and equivalent resistance in the circuit. Therefore rewrite the equation.

$\begin{array}{l}I{R}_{eq}=-\frac{2NBA}{\Delta t}\\ \frac{\Delta q}{\Delta t}{\left(\frac{1}{R_1}+\frac{1}{R_2}\right)}^{-1}=-\frac{2NBA}{\Delta t}\end{array}$ (III)

Simplify equation (III) to get $\Delta q$.

$\begin{array}{l}\Delta q=2NBA\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ =2NB{a}^2\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\end{array}$ (IV)

Conclusion

Substitute $50$ for $N$ , $1.4\text{T}$ for $B$ , $0.45\text{m}$ for $a$ , $10\Omega$ for $R_1$ , $5\Omega$ for $R_2$ in equation (IV) to get $\Delta q$

$\begin{array}{c}\Delta q=2\left(50\right)\left(1.4\text{T}\right){\left(0.45\text{m}\right)}^2\left(\frac{1}{10\Omega }+\frac{1}{5\Omega }\right)\\ =8.5\text{C}\end{array}$

Therefore, The charge that flows through the circuit is $8.5\text{C}$.

(b)

To determine

How much charge goes through the $5.0\Omega$ resistor?

Answer to Problem 96P

The amount of charge flowing through the $5.0\Omega$ resistor is $5.7\text{C}$.

Explanation of Solution

The simplified circuit diagram is shown in figure 1 below. Apply Kirchhoff’s law to the circuit.

Apply junction rule to find the charge flowing through $5\Omega$ resistor.

$\begin{array}{c}I=\frac{\Delta q}{\Delta t}\\ I_1+I_2=\frac{\Delta q_1}{\Delta t}+\frac{\Delta q_2}{\Delta t}\\ \Delta q=\Delta q_1+\Delta q_2\end{array}$ (I)

Here, $I_1$ is the current in $10\Omega$ resistor, $I_2$ is the current in $5\Omega$ resistor, $\Delta q_1$ is the charge flowing in $10\Omega \text{resistor}$ , $\Delta q_2$ is the charge flowing through $5\Omega \text{resistor}$

Apply loop rule to the circuit.

$\begin{array}{l}0=I_1{R}_1-I_2{R}_2\\ =\frac{\Delta q_1}{\Delta t}{R}_1-\frac{\Delta q_1}{\Delta t}{R}_2\end{array}$ (II)

Here, $I_1$ is the current in $10\Omega$ resistor, $I_2$ is the current in $5\Omega$ resistor, $\Delta q_1$ is the charge flowing in $10\Omega \text{resistor}$ , $\Delta q_2$ is the charge flowing through $5\Omega \text{resistor}$

Replace $\Delta q_1$ by $\Delta q-\Delta q_2$ and simplify equation (II) to get $\Delta q_2$

$\begin{array}{c}0=\frac{\Delta q-\Delta q_2}{\Delta t}{R}_1-\frac{\Delta q_2}{\Delta t}{R}_2\\ \Delta q_2=\frac{R_1}{R_1+R_2}\Delta q\end{array}$ (III)

Conclusion:

Substitute $10\Omega$ for $R_1$ , $5\Omega$ for $R_2$ , $8.5\text{C}$ for $\Delta q$ in equation (III) to get $\Delta q_2$

$\begin{array}{c}\Delta q_2=\frac{10\Omega }{10\Omega +5\Omega }\left(8.5\text{C}\right)\\ =5.7\text{C}\end{array}$

Therefore, The amount of charge flowing through the $5.0\Omega$ resistor is $5.7\text{C}$.