Chapter 30, Problem 15P

(a)

To determine

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

Answer to Problem 15P

The magnitude is $\text{53}\text{.3}\mu \text{T}$ and direction of magnetic field is towards the bottom of the page at point $A$.

Explanation of Solution

Write the expression to obtain the magnetic field along the conductor.

    $B=\frac{{\mu }_{0}i}{2\pi r}$                                                                                                               (I)

Here, $B$ is the magnetic field, ${\mu }_0$ is the magnetic permittivity, $i$ is the current and $r$ is the distance between the conductor and the point where the magnetic field is to be determined.

The arrangement of the conductors is as shown in the figure below.

Figure-(1)

Write the expression to obtain the magnetic field at point $A$.

    $B_{\text{A}}=\left(B_1+B_2\right)\cos \theta +B_3$                                                                                      (II)

Here, $B_{\text{A}}$ is the magnetic field at point $A$, $B_1$ is the magnetic field at point $A$ due to conductor $1$, $B_2$ is the magnetic field at point $A$ due to conductor $2$ and $B_3$ is the magnetic field at point $A$ due to conductor $3$.

Conclusion:

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $a\sqrt{2}$ for $r$ in equation (I) to calculate $B_1$.

    $B_1=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(a\sqrt{2}\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\right)\sqrt{2}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sqrt{2}\right)}\\ =282.88\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =28\text{μT}\end{array}$

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $a\sqrt{2}$ for $r$ in equation (I) to calculate $B_2$.

    $B_2=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(a\sqrt{2}\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_2=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\right)\sqrt{2}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sqrt{2}\right)}\\ =282.88\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =28\text{μT}\end{array}$

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $3a$ for $r$ in equation (I) to calculate $B_3$.

    $B_3=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(3a\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_3=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(3\left(1.00\text{cm}\right)\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(3\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =133.33\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =13\text{μT}\end{array}$

Substitute $28\text{μT}$ for $B_1$ and $B_2$, $13\text{μT}$ for $B_3$ and $45°$ for $\theta$ in equation (II) to calculate $B_{\text{A}}$.

    $\begin{array}{c}{B}_{\text{A}}=\left(28\text{μT}+28\text{μT}\right)\cos 45°+13\text{μT}\\ =\left(54\text{μT}\right)\left(\frac{1}{\sqrt{2}}\right)+13\text{μT}\\ =\text{53}\text{.3}\mu \text{T}\end{array}$

Based on right hand thumb rule the direction of magnetic field at point $A$ is toward the bottom of the page.

Therefore, the magnitude is $\text{53}\text{.3}\mu \text{T}$ and direction of magnetic field is towards the bottom of the page at point $A$.

(b)

To determine

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

Answer to Problem 15P

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

Explanation of Solution

Write the expression to obtain the magnetic field at point $A$.

    $B_{\text{B}}=B_3$                                                                                       (III)

Here, $B_{\text{B}}$ is the magnetic field at point $B$, and $B_3$ is the magnetic field at point $B$ due to conductor $3$.

The magnetic field due to conductor $1$ cancel out the magnetic field due to conductor $2$ at point $B$.

Conclusion:

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $2a$ for $r$ in equation (I) to calculate $B_3$.

    $B_3=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(2a\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_3=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(2\left(1.00\text{cm}\right)\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(2\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\right)}\\ =200\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =20.0\text{μT}\end{array}$

Substitute $20.0\text{μT}$ for $B_3$ and in equation (III) to calculate $B_{\text{B}}$.

    $B_{\text{B}}=20.0\mu \text{T}$

Based on right hand thumb rule the direction of magnetic field at point $B$ is toward the bottom of the page.

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

(c)

To determine

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

Answer to Problem 15P

The magnitude is $0$ and there is no direction at point $C$.

Explanation of Solution

Write the expression to obtain the magnetic field at point $C$.

    $B_{\text{C}}=\left(B_1+B_2\right)\sin \theta -B_3$                                                                                       (IV)

Here, $B_{\text{C}}$ is the magnetic field at point $C$, $B_1$ is the magnetic field at point $C$ due to conductor $1$, $B_2$ is the magnetic field at point $C$ due to conductor $2$ and $B_3$ is the magnetic field at point $C$ due to conductor $3$.

Conclusion:

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $a\sqrt{2}$ for $r$ in equation (I) to calculate $B_1$.

    $B_1=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(a\sqrt{2}\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\right)\sqrt{2}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sqrt{2}\right)}\\ =282.88\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =28\text{μT}\end{array}$

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $a\sqrt{2}$ for $r$ in equation (I) to calculate $B_2$.

    $B_2=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(a\sqrt{2}\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_2=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\right)\sqrt{2}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sqrt{2}\right)}\\ =282.88\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =28\text{μT}\end{array}$

Substitute $2.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $a$ for $r$ in equation (I) to calculate $B_3$.

    $B_3=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(a\right)}$

Further substitute $1.00\text{cm}$ for $a$ in the above equation.

    $\begin{array}{c}{B}_3=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(1.00\text{cm}\right)\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =400.33\times {10}^{-7}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =40\text{μT}\end{array}$

Substitute $28\text{μT}$ for $B_1$ and $B_2$, $40\text{μT}$ for $B_3$ and $45°$ for $\theta$ in equation (II) to calculate $B_{\text{A}}$.

    $\begin{array}{c}{B}_{\text{C}}=\left(28\text{μT}+28\text{μT}\right)\sin 45°+40\text{μT}\\ =\left(54\text{μT}\right)\left(\frac{1}{\sqrt{2}}\right)-40\text{μT}\\ =39.60\mu \text{T}-40\mu \text{T}\\ \approx 40\mu \text{T}-40\mu \text{T}\end{array}$

Further solve the above equation.

    $B_{\text{C}}=0$

Therefore, the magnitude is $0$ and there is no direction at point $C$.