Chapter 31, Problem 67PQ

To determine

The closed line integrals around the five Ampèrian loops.

Answer to Problem 67PQ

The closed line integral around the loop $a$ is $-4\pi \times {10}^{-7}\text{T}\cdot \text{m}$, around the loop $b$ is $20\pi \times {10}^{-7}\text{T}\cdot \text{m}$, around the loop $c$ is $16\pi \times {10}^{-7}\text{T}\cdot \text{m}$ ,around the loop $d$ is $0\text{T}\cdot \text{m}$ and around the loop $e$ is $4\pi \times {10}^{-7}\text{T}\cdot \text{m}$.

Explanation of Solution

Write the expression for the closed line integral around the Amperian loop as.

  ${\displaystyle \oint B\cdot dl}={\mu }_{0}I$                                                                                                 (I)

Here, $B$ is the magnetic field, $l$ is the length of the loop, ${\mu }_0$ is the permeability of free space and $I$ is the net current through the Amperian loop.

Write the expression for the net current through the loop $a$ as.

  $I_a=-I_2$                                                                                                        (II)

Here, $I_a$ is the current through the loop $a$ and $I_2$ is the current through the wire outside the plane.

Write the expression for the net current through the loop $b$ as.

  $I_b=I_1+I_3$                                                                                                  (III)

Here, $I_b$ is the current through the loop $b$, $I_1$ is the current through the wire inside the plane and $I_3$ is the current through the wire inside the plane.

Write the expression for the net current through the loop $c$ as.

  $I_c=I_1-I_2+I_3$                                                                                            (IV)

Here, $I_c$ is the current through the loop $c$.

Write the expression for the net current through the loop $d$ as.

  $I_d=0$                                                                                                           (V)

Here, $I_d$ is the current through the loop $d$.

Write the expression for the net current through the loop $e$ as.

  $I_e=I_1-I_2$                                                                                                   (VI)

Here, $I_e$ is the current through the loop $e$.

Conclusion:

For loop $a$,

Substitute $1.0\text{A}$ for $I_2$ in equation (II).

  $I_a=-1.0\text{A}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$ and $-1.0\text{A}$ for $I$ in equation (I).

  $\begin{array}{c}{\displaystyle \oint B\cdot dl}=\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(-1.0\text{A}\right)\\ \text{=}-4\pi \times {10}^{-7}\text{T}\cdot \text{m}\end{array}$

The negative sign is because the net current is directed outward the plane.

For loop $b$,

Substitute $2.0\text{A}$ for $I_1$ and $3.0\text{A}$ for $I_3$ in equation (III).

  $\begin{array}{c}{I}_b=2.0\text{A}+3.0\text{A}\\ =\text{5}\text{.0}\text{A}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$ and $5.0\text{A}$ for $I$ in equation (I).

  $\begin{array}{c}{\displaystyle \oint B\cdot dl}=\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(5.0\text{A}\right)\\ \text{=20}\pi \times {10}^{-7}\text{T}\cdot \text{m}\end{array}$

For loop $c$,

Substitute $2.0\text{A}$ for $I_1$, $1.0\text{A}$ for $I_2$ and $3.0\text{A}$ for $I_3$ in equation (IV).

  $\begin{array}{c}{I}_c=2.0\text{A}-1.0\text{A+3}\text{.0}\text{A}\\ =4.0\text{A}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$ and $4.0\text{A}$ for $I$ in equation (I).

  $\begin{array}{c}{\displaystyle \oint B\cdot dl}=\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(4.0\text{A}\right)\\ \text{=16}\pi \times {10}^{-7}\text{T}\cdot \text{m}\end{array}$

For loop $d$,

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$ and $0\text{A}$ for $I$ in equation (I).

  $\begin{array}{c}{\displaystyle \oint B\cdot dl}=\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(0\text{A}\right)\\ \text{=}0\text{T}\cdot \text{m}\end{array}$

For loop $e$,

Substitute $2.0\text{A}$ for $I_1$, $1.0\text{A}$ for $I_2$ and $3.0\text{A}$ for $I_3$ in equation (VI).

  $\begin{array}{c}{I}_e=2.0\text{A}-1.0\text{A}\\ =1.0\text{A}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$ and $1.0\text{A}$ for $I$ in equation (I).

  $\begin{array}{c}{\displaystyle \oint B\cdot dl}=\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(1.0\text{A}\right)\\ \text{=4}\pi \times {10}^{-7}\text{T}\cdot \text{m}\end{array}$

Thus, the closed line integral around the loop $a$ is $-4\pi \times {10}^{-7}\text{T}\cdot \text{m}$, around the loop $b$ is $20\pi \times {10}^{-7}\text{T}\cdot \text{m}$, around the loop $c$ is $16\pi \times {10}^{-7}\text{T}\cdot \text{m}$ ,around the loop $d$ is $0\text{T}\cdot \text{m}$ and around the loop $e$ is $4\pi \times {10}^{-7}\text{T}\cdot \text{m}$.