Chapter 32, Problem 73PQ

To determine

The amount of charge entering the resistor during the time $\Delta t$.

Answer to Problem 73PQ

The charge entering the resistor during the time $\Delta t$ is $24.43\text{A}$

Explanation of Solution

Write the expression for Faraday’s law of induction.

    $\left|\epsilon \right|=N\left|\frac{d\left({\varphi }_{\text{B}}\right)}{dt}\right|$                                                                                      (I)

Here, $\epsilon$ is the induced emf, $N$ is the number of turns in a loop, $t$ is the time and ${\varphi }_{\text{B}}$ is the magnetic flux through the loop.

Write the equation of magnetic flux through the loop.

    ${\varphi }_{\text{B}}=BA\cos \phi$

Here, $B$ is the magnetic field, $A$ is the area of the loop and $\phi$ is the angle between the magnetic field and area vector.

Substitute $BA\cos \phi$ for ${\varphi }_{\text{B}}$ in equation (I) to find $\left|\epsilon \right|$.

    $\begin{array}{c}\left|\epsilon \right|=N\left|\frac{d\left(BA\cos \phi \right)}{dt}\right|\\ =NBA\left|\frac{d\left(\cos \phi \right)}{dt}\right|\\ =NBA\left|\frac{\cos {\phi }_2-\cos {\phi }_1}{\Delta t}\right|\end{array}$                                                                             (II)

Write the formula to calculate the area of the circular coil.

    $A=\pi r^2$

Here, $r$ is the radius of the coil.

Substitute $\pi r^2$ for $A$ in equation (II) to find $\left|\epsilon \right|$.

    $\left|\epsilon \right|=NB\pi r^2\left|\frac{\cos {\phi }_2-\cos {\phi }_1}{\Delta t}\right|$                                                                   (III)

Write the expression for Ohm’s law.

    $\begin{array}{c}\epsilon =IR\\ I\text{=}\frac{\epsilon }{R}\end{array}$                                                                                                   (IV)

Here, $R$ is the resistance of the coil, $I$ is the current in the coil.

Write the expression to calculate the charge.

    $q=I\Delta t$                                                                                                (V)

Here, $q$ is the charge.

Conclusion:

Substitute $320$ for $N$, $3\text{T}$ for $B$, $9\text{cm}$ for $r$, $180°$ for ${\phi }_2$ and $0°$ for ${\phi }_1$ in equation (III) to find $\left|\epsilon \right|$.

    $\begin{array}{c}\left|\epsilon \right|=\left|320\times 3\text{T}\times 3.142\times \right|{\left(9\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2\left|\frac{\cos 180°-\cos 0°}{\Delta t}\right|\\ =\frac{48.86\text{V}}{\Delta t}\end{array}$

Substitute $\frac{48.86\text{V}}{\Delta t}$ for $\epsilon$ and $2.4\Omega$ for $R$ in equation (IV) to find $I$.

    $\begin{array}{c}I=\frac{\left(\frac{48.86\text{V}}{\Delta t}\right)}{2.4\Omega }\\ =\frac{24.43\text{A}}{\Delta t}\end{array}$

Substitute $\frac{24.43\text{A}}{\Delta t}$ for $I$ in equation (V) to find $q$.

    $\begin{array}{c}q=\frac{24.43\text{A}}{\Delta t}\times \Delta t\\ =24.43\text{A}\end{array}$

Therefore, the charge entering the resistor during the time $\Delta t$ is $24.43\text{A}$.