Chapter 19, Problem 39E

(a)

To determine

The magnitude and the direction of the magnetic field at the origin.

Answer to Problem 39E

The magnitude of the magnetic field at the origin due to two parallel current carrying wire is ${10}^{-5}\text{T}$ and the magnitude is out of page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

    $B=\frac{{\mu }_{0}I}{2\pi d}$        (1)

Here, ${\mu }_0$ is the magnetic permeability in free space, $I$ is the current conducted by first wire, and $d$ is the distance of a point from the wire.

Write the expression for the net magnetic field due to two parallel conducting wire.

    $B=\left(B_1-B_2\right)$        (2)

Here, $B_1$ is the field due to first wire and $B_2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B_1$ for $B$, $\left(4\pi \times {10}^{-7}\right)$ for ${\mu }_0$, $10\text{A}$ for $I$ and $0.1\text{m}$ for $d$ in equation (1).

    $\begin{array}{c}{B}_1=\frac{ 4\pi \times {10}^{-7}\left(10\text{A}\right)}{2\pi \left(0.1\text{m}\right)}\\ =2\times {10}^{-5}\text{T}\end{array}$

The magnetic field due to the second wire is calculated below.

Substitute $B_1$ for $B$, $\left(4\pi \times {10}^{-7}\right)$ for ${\mu }_0$, $5\text{A}$ for $I$ and $0.1\text{m}$ for $d$ in equation (1).

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\text{A}\right)}{2\pi \left(0.1\text{m}\right)}\\ ={10}^{-5}\text{T}\end{array}$

Substitute $2\times {10}^{-5}\text{T}$ for $B_1$ and ${10}^{-5}\text{T}$ for $B_2$ in equation (2).

    $\begin{array}{c}B=\left(2\times {10}^{-5}\text{T}-{10}^{-5}\text{T}\right)\\ ={10}^{-5}\text{T}\end{array}$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is ${10}^{-5}\text{T}$ and the magnitude is out of page.

(b)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$-axis.

Answer to Problem 39E

The magnitude of the magnetic field at any point in the positive $x$-axis due to two parallel current carrying wire is $1.67\times {10}^{-5}\text{T}$ and the magnitude is in page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

    $B=\frac{{\mu }_{0}I}{2\pi d}$

Write the expression for the net magnetic field due to two parallel conducting wire.

    $B=-\left(B_1+B_2\right)$        (3)

Here, $B_1$ is the field due to first wire and $B_2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B_1$ for $B$ and $\left(x-a\right)$ for $d$ in equation (1).

    $B_1=\frac{{\mu }_{0}I}{2\pi \left(x-a\right)}$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute  $\left(4\pi \times {10}^{-7}\right)$ for ${\mu }_0$, $5\text{A}$ for $I$ and $0.1\text{m}$ for $a$ and $0.2\text{m}$ for $x$ in the above equation.

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\text{A}\right)}{2\pi \left(0.2\text{m}-0.1\text{m}\right)}\\ ={10}^{-5}\text{T}\end{array}$

The magnetic field due to the second wire is calculated below.

Substitute $B_2$ for $B$, $I^{\prime }$ for $I$ and $\left(a+x\right)$ for $d$ in equation (1).

    $B_2=\frac{{\mu }_0{I}^{\prime }}{2\pi \left(a+x\right)}$

Substitute $\left(4\pi \times {10}^{-7}\right)$ for ${\mu }_0$, $\text{10}\text{A}$ for $I$ and $0.1\text{m}$ for $a$ and $0.2\text{m}$ for $x$ in the above equation.

    $\begin{array}{c}{B}_2=\frac{\left(4\pi \times {10}^{-7}\right)\left(10\text{A}\right)}{2\pi \left(0.2\text{m+}0.1\text{m}\right)}\\ =6.67\times {10}^{-6}\text{T}\end{array}$

Substitute ${10}^{-5}\text{T}$ for $B_1$ and $6.67\times {10}^{-6}\text{T}$ for $B_2$ in equation (3).

    $\begin{array}{c}B=-\left({10}^{-5}\text{T+}6.67\times {10}^{-6}\text{T}\right)\\ =-1.67\times {10}^{-5}\text{T}\end{array}$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $1.67\times {10}^{-5}\text{T}$ and the magnitude is in page.

(c)

To determine

The magnitude and the direction of the magnetic field at any point in the positive $x$-axis.

Answer to Problem 39E

The magnitude of the magnetic field at any point in the negative $x$-axis due to two parallel current carrying wire is $2.33\times {10}^{-5}\text{T}$ and the magnitude is out of the page.

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

    $B=\frac{{\mu }_{0}I}{2\pi d}$

Write the expression for the net magnetic field due to two parallel conducting wire.

    $B=\left(B_1+B_2\right)$        (4)

Here, $B_1$ is the field due to first wire and $B_2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to the first wire is calculated below.

Substitute $B_1$ for $B$ and $\left(a+x\right)$ for $d$ in equation (1).

    $B_1=\frac{{\mu }_{0}I}{2\pi \left(a+x\right)}$

Here, $a$ is the distance of the wire from the origin and $a$ is the distance of the point from the origin.

Substitute  $\left(4\pi \times {10}^{-7}\right)$ for ${\mu }_0$, $5\text{A}$ for $I$ and $0.1\text{m}$ for $a$ and $0.2\text{m}$ for $x$ in the above equation.

    $\begin{array}{c}{B}_1=\frac{\left(4\pi \times {10}^{-7}\right)\left(5\text{A}\right)}{2\pi \left(0.2\text{m+}0.1\text{m}\right)}\\ =3.33\times {10}^{-6}\text{T}\end{array}$

The magnetic field due to the second wire is calculated below.

Substitute $B_2$ for $B$, $I^{\prime }$ for $I$ and $\left(x-a\right)$ for $d$ in equation (1).

    $B_2=\frac{{\mu }_0{I}^{\prime }}{2\pi \left(x-a\right)}$

Substitute $\left(4\pi \times {10}^{-7}\right)$ for ${\mu }_0$, $\text{10}\text{A}$ for $I$ and $0.1\text{m}$ for $a$ and $0.2\text{m}$ for $x$ in the above equation.

    $\begin{array}{c}{B}_2=\frac{\left(4\pi \times {10}^{-7}\right)\left(10\text{A}\right)}{2\pi \left(0.2\text{m}-0.1\text{m}\right)}\\ =2\times {10}^{-5}\text{T}\end{array}$

Substitute $3.33\times {10}^{-6}\text{T}$ for $B_1$ and $2\times {10}^{-5}\text{T}$ for $B_2$ in equation (4).

    $\begin{array}{c}B=\left(3.33\times {10}^{-6}\text{T+}2\times {10}^{-5}\text{T}\right)\\ =2.33\times {10}^{-5}\text{T}\end{array}$

Thus, the magnitude of the magnetic field at the origin due to two parallel current carrying wire is $2.33\times {10}^{-5}\text{T}$ and the magnitude is out of the page.

(d)

To determine

The value of $x$ at which the net magnetic field due to two current carrying wire is $0$.

Answer to Problem 39E

The distance is $0.0333\text{m}$ at which the magnetic field is zero

Explanation of Solution

Write the expression for the magnetic field due to a conducting wire.

    $B=\frac{{\mu }_{0}I}{2\pi d}$

Write the expression for the net magnetic field due to two parallel conducting wire.

    $B=\left(B_1+B_2\right)$        (5)

Here, $B_1$ is the field due to first wire and $B_2$ is the magnetic field due to second wire.

Conclusion:

The magnetic field due to wire at positive $x$-axis is calculated below.

Substitute $B_1$ for $B$ and $\left(a-x\right)$ for $d$ in equation (1).

    $B_1=\frac{{\mu }_{0}I}{2\pi \left(a-x\right)}$

Here, $a$ is the distance of the wire from the origin and $x$ is the distance of the point from the origin.

The magnetic field due to wire at negative $x$-axis is calculated below.

Substitute $B_2$ for $B$, $I^{\prime }$ for $I$ and $\left(a+x\right)$ for $d$ in equation (1).

    $B_2=\frac{{\mu }_0{I}^{\prime }}{2\pi \left(a+x\right)}$

Substitute $\frac{{\mu }_{0}I}{2\pi \left(a-x\right)}$ for $B_1$ and $\frac{{\mu }_0{I}^{\prime }}{2\pi \left(a+x\right)}$ for $B_2$ in equation (5).

    $B=\frac{{\mu }_{0}I}{2\pi \left(a-x\right)}-\frac{{\mu }_0{I}^{\prime }}{2\pi \left(a+x\right)}$

Substitute $0$ for $B$ and rearrange the above equation.

    $\begin{array}{l}0=\frac{{\mu }_{0}I}{2\pi \left(a-x\right)}-\frac{{\mu }_0{I}^{\prime }}{2\pi \left(a+x\right)}\\ \frac{I}{\left(a-x\right)}=\frac{I^{\prime }}{\left(a+x\right)}\end{array}$

Substitute $10\text{A}$ for $I$, $5\text{A}$ for $I^{\prime }$ and $0.1\text{m}$ for $a$ in the above equation.

    $\begin{array}{c}\frac{10\text{A}}{\left(0.1\text{m}-x\right)}=\frac{5\text{A}}{\left(0.1\text{m}+x\right)}\\ 2\left(x+0.1\text{m}\right)=\left(0.1\text{m}-x\right)\end{array}$

Simplify the above equation.

    $x\approx 0.0333\text{m}$

Thus, the distance is $0.0333\text{m}$ at which the magnetic field is zero.