Chapter 22, Problem 37P

(a)

To determine

The magnitude and direction of magnetic field at the point $A$

Answer to Problem 37P

The direction of the magnetic field is towards the bottom of the page and magnitude of magnetic field at the point $A$ is $53.3\text{T}$.

Explanation of Solution

From the figure 1

Write the expression to calculate the side of the square,

  $\begin{array}{c}x=\sqrt{a^2+a^2}\\ =\sqrt{2}a\end{array}$        (I)

Write the formula to calculate the angle $\theta$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{a}{a}\right)\\ =45°\end{array}$                                                                                              (II)

Figure.2 shows the direction of the magnetic fields ${\overrightarrow{B}}_1$, ${\overrightarrow{B}}_2$ and ${\overrightarrow{B}}_3$. The horizontal components of ${\overrightarrow{B}}_1$ and ${\overrightarrow{B}}_2$ cancel each other and their vertical components add together to give ${\overrightarrow{B}}_3$.

Write the expression for magnetic field at point $A$ due to conductor $1$ ,

  $B_{A1}=\frac{{\mu }_{0}I}{2\pi r}$        (III)

Here, $I$ is the current flowing through conductor, $r$ is the distance of the point $A$ due to conductor 1.

Substitute $\sqrt{2}a$ for $r$ in the above expression (III)

  $B_{A1}=\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}$        (IV)

Write the expression for magnetic field at point $A$ due to conductor $2$ ,

  $B_{A2}=\frac{{\mu }_{0}I}{2\pi r}$        (V)

Here, $I$ is the current flowing through conductor, $r$ is the distance of the point $A$ due to conductor 2.

Substitute $\sqrt{2}a$ for $r$ in the above expression (V)

  $B_{A2}=\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}$        (VI)

Write the expression for magnetic field at point $A$ due to conductor $3$ ,

  $B_{A3}=\frac{{\mu }_{0}I}{2\pi r}$        (VII)

Here, $I$ is the current flowing through conductor, $r$ is the distance of the point $A$ due to conductor $3$.

Substitute $3a$ for $r$ in the above expression (VII)

  $B_{A3}=\frac{{\mu }_{0}I}{6\pi a}$        (VIII)

Write the expression to calculate magnetic field at the point $A$ due to the conductors in $x$ direction.

  $\begin{array}{c}{\displaystyle \sum B_x=0}\\ B_{Ax}-B_{A3}+B_{A2}\cos \theta +B_{A1}\cos \theta =0\\ B_{Ax}=B_{A3}+B_{A2}\cos \theta +B_{A1}\cos \theta \end{array}$        (X)

Substitute $\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}$ for $B_{A1}$  and $B_{A2}$, and  $\frac{{\mu }_{0}I}{6\pi a}$ for $B_{A3}$ and $45°$ for $\theta$  in expression (IX)

  $\begin{array}{c}{B}_{Ax}=\frac{{\mu }_{0}I}{6\pi a}+\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}\cos 45°+\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}\cos 45°\\ =\frac{{\mu }_{0}I}{6\pi a}+\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}\left(\frac{1}{\sqrt{2}}\right)+\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{{\mu }_{0}I}{6\pi a}+\frac{{\mu }_{0}I}{4\pi a}+\frac{{\mu }_{0}I}{4\pi a}\\ =\frac{2}{3}\frac{{\mu }_{0}I}{\pi a}\end{array}$        (X)

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ , $2.00\text{A}$ for $I$ , and $1.00\text{cm}$ for $a$ in the expression (X),

    $\begin{array}{c}{B}_{Ax}=\frac{2}{3}\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(2.00\text{A}\right)}{\pi \left(1.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =53.3\times {10}^{-6}\text{T}\\ \text{=53}\text{.3}\mu \text{T}\end{array}$

Write the expression to calculate the $y$ -component of magnetic field.

    $\begin{array}{c}{\displaystyle \sum B_y=0}\\ B_{Ay}-B_{A2}\cos 45°+B_{A1}\cos 45°=0\end{array}$        (XI)

Substitute $\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}$ for $B_{A1}$  and $B_{A2}$  in expression (XI)

    $\begin{array}{c}{B}_{Ay}-\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}\cos 45°+\frac{{\mu }_{0}I}{2\sqrt{2}\pi a}\cos 45°=0\\ B_{Ay}=0\end{array}$

Write the expression to calculate the net magnetic field at point $A$ ,

    $B_A=\sqrt{B_{Ax}{}^2+B_{Ay}{}^2}$        (XII)

Substitute $\text{53}\text{.3}\mu \text{T}$ for $B_{Ax}$ and $0$ for $B_{Ay}$ in the expression (XII),

    $\begin{array}{c}{B}_A=\sqrt{{\left(\text{53}\text{.3}\mu \text{T}\right)}^2+0}\\ =\text{53}\text{.3}\mu \text{T}\end{array}$

The direction of the magnetic field is towards the bottom of the page and magnitude of magnetic field at the point $A$ is $53.3\text{T}$.

(b)

To determine

The magnitude and direction of the magnetic field at the point $B$ .

Answer to Problem 37P

The magnitude of magnetic field at point $B$ is $20.0\text{μT}$ and the direction is toward the bottom of the page.

Explanation of Solution

The sum of magnetic fields due to conductors 1 and 2 at point $B$ is zero because they have equal magnitudes and opposite directions. The net magnetic field at that point $B$ is due to the conductor 3.

Write the relation for the net magnetic field at point $B$ is,

$B_B=\frac{{\mu }_{0}I}{4\pi a}$

Calculation:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ , $2.00\text{A}$ for $I$ , and $1.00\text{cm}$ for $a$ b to find $B_B$

$\begin{array}{l}{B}_B=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)\left(2.00\text{A}\right)}{4\pi a\left(1.00\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1.00\text{cm}}\right)}\\ =20.0\text{μT}\text{toward the bottom of the page}\end{array}$

Therefore, the magnitude of magnetic field at point $B$ is $20.0\text{μT}$ and the direction is toward the bottom of the page.

(c)

To determine

The magnitude and direction of the magnetic field at the point $C$ .

Answer to Problem 37P

The net magnitude of the magnetic field at the point $C=0$ .

Explanation of Solution

Figure.3 shows the direction of the magnetic fields ${\overrightarrow{B}}_1$, ${\overrightarrow{B}}_2$ and ${\overrightarrow{B}}_3$. Here also, the horizontal components of ${\overrightarrow{B}}_1$ and ${\overrightarrow{B}}_2$ cancel each other and their vertical components add together to give ${\overrightarrow{B}}_3$.

The net magnetic field at point $C$ is,

$\begin{array}{c}{B}_C=B_1+B_2-B_3\\ =2B_1-B_3\\ =2\left[\frac{{\mu }_{0}I}{2\pi \left(a\sqrt{2}\right)}\cos 45.0°\right]-\frac{{\mu }_{0}I}{2\pi a}\\ =0\end{array}$

Conclusion:

Therefore, the net magnetic field at the point $C$ is $0$.