Chapter 30, Problem 77PQ

(a)

To determine

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

Answer to Problem 77PQ

The magnetic field at point $A$ is $3.13\times {10}^{-6}\widehat{k}\text{T}$.

Explanation of Solution

The direction of the magnetic field at the point $A$ is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current $I_1$ as.

${\overrightarrow{B}}_{1A}=\frac{{\mu }_0{I}_1}{2\pi b}\widehat{k}$ (I)

Here, ${\overrightarrow{B}}_{1A}$ is the magnetic field due to current $I_1$, ${\mu }_0$ is the permeability of free space, $I_1$ is the current flowing in the wire and $b$ is the distance of point $A$ from the wire.

Write the expression for the magnetic field due to wire carrying current $I_2$ as.

${\overrightarrow{B}}_{2A}=\frac{{\mu }_0{I}_2}{2\pi a}\widehat{k}$ (II)

Here, ${\overrightarrow{B}}_{2A}$ is the magnetic field due to current $I_2$ , ${\mu }_0$ is the permeability of free space, $I_2$ is the current flowing in the wire and $a$ is the distance of point $A$ from the wire.

Write the expression for the net magnetic field as.

  ${\overrightarrow{B}}_A={\overrightarrow{B}}_{1A}+{\overrightarrow{B}}_{2A}$                                                                                             (III)

Here, ${\overrightarrow{B}}_A$ is the net magnetic field at point $A$.

Conclusion:

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$$2.00\text{A}$ for $I_1$ and $50.0\text{cm}$ for $b$ in equation (I).

  $\begin{array}{c}{\overrightarrow{B}}_{1A}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(2.00\text{A}\right)}{2\pi \left(50.0\text{cm}\right)}\widehat{k}\\ =\frac{4\times {10}^{-7}\text{T}\cdot \text{m}}{\left(50.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\widehat{k}\\ =\left(8\times {10}^{-7}\text{T}\right)\widehat{k}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$$7.00\text{A}$ for $I_2$ and $60.0\text{cm}$ for $a$ in equation (II).

  $\begin{array}{c}{\overrightarrow{B}}_{2A}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(7.00\text{A}\right)}{2\pi \left(60.0\text{cm}\right)}\widehat{k}\\ =\frac{14\times {10}^{-7}\text{T}\cdot \text{m}}{\left(60.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\widehat{k}\\ =\left(2.33\times {10}^{-6}\text{T}\right)\widehat{k}\end{array}$

Substitute $8\times {10}^{-7}\widehat{k}\text{T}$ for ${\overrightarrow{B}}_{1A}$ and $2.33\times {10}^{-6}\widehat{k}\text{T}$ for ${\overrightarrow{B}}_{2A}$ in equation (III).

  $\begin{array}{c}{\overrightarrow{B}}_A=\left(8\times {10}^{-7}\text{T}\right)\widehat{k}+\left(2.33\times {10}^{-6}\text{T}\right)\widehat{k}\\ =\left(3.13\times {10}^{-6}\text{T}\right)\widehat{k}\end{array}$

Thus, the magnetic field at point $A$ is $3.13\times {10}^{-6}\widehat{k}\text{T}$.

(b)

To determine

The magnitude and direction of magnetic field at point $z=-50\text{cm}$.

Answer to Problem 77PQ

The magnetic field at point $z=-50\text{cm}$ is $\left(2.80\times {10}^{-6}\text{T}\right)\widehat{i}-\left(0.80\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)$.

Explanation of Solution

Consider the point at $z=-50\text{cm}$ as point $B$.

The direction of the magnetic field at the point $B$ is given by the Right Hand Rule. The thumb of the right hand is kept in the direction of the current and the palm faces the point, then the curl of the fingers give the direction of the magnetic field.

Write the expression for the magnetic field due to wire carrying current $I_1$ as.

${\overrightarrow{B}}_{1B}=\frac{{\mu }_0{I}_1}{2\pi r}\left(-\widehat{j}\right)$ (IV)

Here, ${\overrightarrow{B}}_{1B}$ is the magnetic field due to current $I_1$ , ${\mu }_0$ is the permeability of free space, $I_1$ is the current flowing in the wire and $r$ is the distance of point $B$ from the wire.

Write the expression for the magnetic field due to wire carrying current $I_2$ as.

${\overrightarrow{B}}_{2B}=\frac{{\mu }_0{I}_2}{2\pi r}\widehat{i}$ (V)

Here, ${\overrightarrow{B}}_{2B}$ is the magnetic field due to current $I_2$ , ${\mu }_0$ is the permeability of free space and $I_2$ is the current flowing in the wire.

Write the expression for the net magnetic field as.

  ${\overrightarrow{B}}_B={\overrightarrow{B}}_{1B}+{\overrightarrow{B}}_{2B}$                                                                                             (VI)

Here, ${\overrightarrow{B}}_B$ is the net magnetic field at point $B$.

Conclusion:

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$$2.00\text{A}$ for $I_1$ and $50.0\text{cm}$ for $r$ in equation (IV).

  $\begin{array}{c}{\overrightarrow{B}}_{1B}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(2.00\text{A}\right)}{2\pi \left(50.0\text{cm}\right)}\left(-\widehat{j}\right)\\ =\frac{4\times {10}^{-7}\text{T}\cdot \text{m}}{\left(50.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\left(-\widehat{j}\right)\\ =\left(8\times {10}^{-7}\text{T}\right)\left(-\widehat{j}\right)\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}$ for ${\mu }_0$$7.00\text{A}$ for $I_2$ and $50.0\text{cm}$ for $r$ in equation (V).

  $\begin{array}{c}{\overrightarrow{B}}_{2B}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{T}\cdot \text{m}}{\text{A}}\right)\left(7.00\text{A}\right)}{2\pi \left(50.0\text{cm}\right)}\widehat{i}\\ =\frac{14\times {10}^{-7}\text{T}\cdot \text{m}}{\left(50.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}\widehat{i}\\ =\left(2.80\times {10}^{-6}\text{T}\right)\widehat{i}\end{array}$

Substitute $-8\times {10}^{-7}\widehat{j}\text{T}$ for ${\overrightarrow{B}}_{1B}$ and $2.80\times {10}^{-6}\widehat{i}\text{T}$ for ${\overrightarrow{B}}_{2B}$ in equation (III).

  $\begin{array}{c}{\overrightarrow{B}}_B=\left(8\times {10}^{-7}\text{T}\right)\left(-\widehat{j}\right)+\left(2.80\times {10}^{-6}\text{T}\right)\widehat{i}\\ =\left(2.80\times {10}^{-6}\text{T}\right)\widehat{i}-\left(0.80\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)\end{array}$

Thus, the magnetic field at point $z=-50\text{cm}$ is $\left(2.80\times {10}^{-6}\text{T}\right)\widehat{i}-\left(0.80\times {10}^{-6}\text{T}\right)\left(\widehat{j}\right)$.