Chapter 31, Problem 31.61AP

The circuit in Figure P3 1.61 is located in a magnetic field whose magnitude varies with lime according to the expression B = 1.00 × 10^-3 t, where B is in teslas and f is in seconds. Assume the resistance per length of the wire is 0.100 Ω/m. Find the current in section PQ of length a = 65.0 cm.

To determine

The current in the section PQ.

Answer to Problem 31.61AP

The current in the section PQ is $283\text{μA}$ upward.

Explanation of Solution

Given Info: The time varying magnetic field is $1.00\times {10}^{-3}t$, the resistance per unit length is $0.100\Omega /\text{m}$ and the length $a$ is $65.0\text{cm}$.

The circuit diagram is as shown below.

Figure (1)

For loop 1:

The resistance of the loop is,

$\begin{array}{c}R=r(2a)+r(2a)+r(a)+R_{\text{PQ}}\\ =\left(r\right)5a+R_{\text{PQ}}\end{array}$

Here,

$r$ is the resistance per unit length.

The length of PQ is $a$, then the resistance of PQ is,

$R_{\text{PQ}}=ra$

The area of the loop is,

$\begin{array}{c}A=\left(2a\right)\left(a\right)\\ =2a^2\end{array}$

Thus, the area of the loop 1 is $2a^2$.

The flux induced in the loop is,

$\varphi =BA\cos \theta$

Here,

$B$ is the magnetic field in the loop.

$A$ is the area of the loop.

$\theta$ is the angle between the normal component of the area and the magnetic field.

The angle between the normal component of the area and the magnetic field is,

$\theta =0°$.

Substitute $0°$ for $\theta$ and $2a^2$ for $A$ and $1.00\times {10}^{-3}t$ for $B$ in the above equation.

$\begin{array}{c}\varphi =1.00\times {10}^{-3}t\left(2a^2\right)\cos \left(0°\right)\\ =1.00\times {10}^{-3}t\left(2a^2\right)\end{array}$

The emf induced in the loop is,

${\left|\epsilon \right|}_1=\frac{d}{dt}\left(\varphi \right)$

Substitute $1.00\times {10}^{-3}t\left(2a^2\right)$ for $\varphi$ in the above equation.

$\begin{array}{c}{\left|\epsilon \right|}_1=\frac{d}{dt}\left(1.00\times {10}^{-3}t\left(2a^2\right)\right)\\ =\left(1.00\times {10}^{-3}\left(2a^2\right)\right)\end{array}$

Thus, the induced emf in the loop 1 is $\left(1.00\times {10}^{-3}\left(2a^2\right)\right)$.

The current in the loop is $I_1$ and current in the PQ is $I_{\text{PQ}}$.

Apply Kirchhoff’s loop rule in loop 1.

${\left|\epsilon \right|}_1-I_1\left(5ar\right)-I_{\text{PQ}}{R}_{\text{PQ}}=0$

Substitute $ra$ for $R_{\text{PQ}}$ in the above equation.

${\left|\epsilon \right|}_1-I_1\left(5ar\right)-I_{\text{PQ}}\left(ar\right)=0$ (1)

For loop (2):

The resistance of the loop is,

$\begin{array}{c}R=r(a)+r(a)+r(a)+R_{\text{PQ}}\\ =\left(r\right)3a+R_{\text{PQ}}\end{array}$

The area of the loop is,

$\begin{array}{c}A=\left(a\right)\left(a\right)\\ =a^2\end{array}$

Thus, the area of the loop 2 is $a^2$.

The flux induced in the loop is,

$\varphi =BA\cos \theta$

The angle between the normal component of the area and the magnetic field is,

$\theta =0°$.

Substitute $0°$ for $\theta$ and $a^2$ for $A$ and $1.00\times {10}^{-3}t$ for $B$ in the above equation.

$\begin{array}{c}\varphi =1.00\times {10}^{-3}t\left(a^2\right)\cos \left(0°\right)\\ =1.00\times {10}^{-3}t\left(a^2\right)\end{array}$

Thus, the flux induced in the loop is $1.00\times {10}^{-3}t\left(a^2\right)$.

The emf induced in the loop is,

${\left|\epsilon \right|}_2=\frac{d}{dt}\left(\varphi \right)$

Substitute $1.00\times {10}^{-3}t\left(a^2\right)$ for $\varphi$ in the above equation.

$\begin{array}{c}{\left|\epsilon \right|}_2=\frac{d}{dt}\left(1.00\times {10}^{-3}t\left(a^2\right)\right)\\ =\left(1.00\times {10}^{-3}\left(a^2\right)\right)\end{array}$

Thus, the induced emf in the loop 2 is $\left(1.00\times {10}^{-3}\left(a^2\right)\right)$.

The current in the loop is $I_2$ and current in the PQ is $I_{\text{PQ}}$.

Apply Kirchhoff’s loop rule in loop 2.

${\left|\epsilon \right|}_2-I_2\left(3ar\right)+I_{\text{PQ}}{R}_{\text{PQ}}=0$

Substitute $ra$ for $R_{\text{PQ}}$ in the above equation.

${\left|\epsilon \right|}_2-I_2\left(3ar\right)+I_{\text{PQ}}\left(ar\right)=0$ (2)

From the figure (1) the current in the arm PQ is,

$\begin{array}{l}{I}_{\text{PQ}}=I_1-I_2\\ I_1=I_{\text{PQ}}+I_2\end{array}$

Substitute $I_{\text{PQ}}+I_2$ for $I_1$ in equation (1).

$\begin{array}{c}{\left|\epsilon \right|}_1-\left(I_{\text{PQ}}+I_2\right)\left(5ar\right)-I_{\text{PQ}}\left(ar\right)=0\\ {\left|\epsilon \right|}_1-\left(5ar\right)I_{\text{PQ}}-\left(5ar\right)I_2-I_{\text{PQ}}\left(ar\right)=0\end{array}$ (3)

Rearrange the equation (2) for $I_2$.

$I_2=\frac{1}{3ar}\left({\left|\epsilon \right|}_2+I_{\text{PQ}}\left(ar\right)\right)$

Substitute $\frac{1}{3ar}\left({\left|\epsilon \right|}_2+I_{\text{PQ}}\left(ar\right)\right)$ for $I_2$ in the equation (3).

$\begin{array}{c}{\left|\epsilon \right|}_1-\left(5ar\right)I_{\text{PQ}}-\left(5ar\right)\left(\frac{1}{3ar}\left({\left|\epsilon \right|}_2+I_{\text{PQ}}\left(ar\right)\right)\right)-I_{\text{PQ}}\left(ar\right)=0\\ {\left|\epsilon \right|}_1-\left(5ar\right)I_{\text{PQ}}-\frac{5}{3}\left({\left|\epsilon \right|}_2+I_{\text{PQ}}\left(ar\right)\right)-I_{\text{PQ}}\left(ar\right)=0\\ {\left|\epsilon \right|}_1-\left(6ar\right)I_{\text{PQ}}-\frac{5}{3}\left({\left|\epsilon \right|}_2+I_{\text{PQ}}\left(ar\right)\right)=0\end{array}$

Substitute $\left(1.00\times {10}^{-3}\left(a^2\right)\right)$ for ${\left|\epsilon \right|}_2$ and $\left(1.00\times {10}^{-3}\left(2a^2\right)\right)$ for ${\left|\epsilon \right|}_1$ in the above equation.

$\left(1.00\times {10}^{-3}\left(2a^2\right)\right)-\left(6ar\right)I_{\text{PQ}}-\frac{5}{3}\left(\left(\left(1.00\times {10}^{-3}\left(a^2\right)\right)\right)+I_{\text{PQ}}\left(ar\right)\right)=0$

Rearrange the above equation for $I_{\text{PQ}}$.

$\begin{array}{c}\left(1.00\times {10}^{-3}\left(2a^2\right)\right)-\left(6ar\right)I_{\text{PQ}}=\frac{5}{3}\left(\left(1.00\times {10}^{-3}\left(a^2\right)\right)+I_{\text{PQ}}\left(ar\right)\right)\\ I_{\text{PQ}}\left(6ar+\frac{5}{3ar}\right)=\left(1.00\times {10}^{-3}\left(2a^2\right)\right)-\frac{5}{3}\left(\left(1.00\times {10}^{-3}\left(a^2\right)\right)\right)\\ I_{\text{PQ}}=\frac{\left(1.00\times {10}^{-3}\left(a^2\right)\right)}{23ar}\end{array}$

Substitute $0.100\Omega /\text{m}$ for $r$ and $65.0\text{cm}$ for $a$ in the above equation.

$\begin{array}{c}{I}_{\text{PQ}}=\frac{\left(1.00\times {10}^{-3}{\left(65.0\text{cm}\right)}^2\right)}{23\left(65.0\text{cm}\right)\left(0.100\Omega /\text{m}\right)}\\ =\frac{\left(1.00\times {10}^{-3}{\left(65.0\text{cm}\times \frac{\text{1}\text{.00}\text{m}}{100\text{cm}}\right)}^2\right)}{23\left(65.0\text{cm}\times \frac{\text{1}\text{.00}\text{m}}{100\text{cm}}\right)\left(0.100\Omega /\text{m}\right)}\\ =\frac{\left(1.00\times {10}^{-3}\left(0.422\text{m}\right)\right)}{23\left(0.0650\Omega \right)}\\ =282.28\times {10}^{-6}\text{A}\end{array}$

Further solve the above equation.

$\begin{array}{c}{I}_{\text{PQ}}=282.28\times {10}^{-6}\text{A}\left(\frac{1\text{μA}}{{10}^{-6}\text{A}}\right)\\ =282.28\text{μA}\\ \approx \text{283}\text{μA}\end{array}$

The current in PQ is $I_1-I_2$ and the value is positive thus it is the direction of $I_1$ hence its is upwards.

Conclusion:

Therefore, the current in the PQ arm is $\text{283}\text{μA}$ upwards.