Chapter 30, Problem 19P

(a)

To determine

The magnitude and the direction of the net magnetic field at the mid way between the wires.

Answer to Problem 19P

The magnitude of the net magnetic field is $-40.0\mu \text{T}$ and the direction is into the page at the mid way between the wires.

Explanation of Solution

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

    $B=\frac{{\mu }_{0}i}{2\pi r}$

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.

Write the expression to obtain the magnetic field due to the wire $1$ at the midway between the wires.

    $B_1=\frac{{\mu }_{0}i}{2\pi r_1}$

Here, $B_1$ is the magnetic field due to the wire $1$ at the midway between the wires and $r_1$ is the distance between the wire $1$ and the point where magnetic field is to be determined.

Write the expression to obtain the magnetic field due to the wire $2$ at the midway between the wires.

    $B_2=\frac{{\mu }_{0}i}{2\pi r_2}$

Here, $B_2$ is the magnetic field due to the wire $2$ at the midway between the wires and $r_2$ is the distance between the wire $2$ and the point where magnetic field is to be determined.

Write the expression to obtain the net magnetic field at the mid way between the wires.

    $B_{\text{net}}=-B_1-B_2$

Here, $B_{\text{net}}$ is the net magnetic field at the mid way between the wires.

Substitute $\frac{{\mu }_{0}i}{2\pi r_1}$ for $B_1$ and $\frac{{\mu }_{0}i}{2\pi r_2}$ for $B_2$ in the above equation.

    $\begin{array}{c}{B}_{\text{net}}=-\frac{{\mu }_{0}i}{2\pi r_1}-\frac{{\mu }_{0}i}{2\pi r_2}\\ =-\frac{{\mu }_{0}i}{2\pi }\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\end{array}$

Further substitute $\frac{d}{2}$ for $r_1$ and $r_2$ in the above equation.

    $\begin{array}{c}{B}_{\text{net}}=-\frac{{\mu }_{0}i}{2\pi }\left(\frac{1}{\frac{d}{2}}+\frac{1}{\frac{d}{2}}\right)\\ =-\frac{{\mu }_{0}i}{2\pi }\left(\frac{2}{d}+\frac{2}{d}\right)\\ =-\frac{2{\mu }_{0}i}{\pi d}\end{array}$

Conclusion:

Substitute $5.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_0$ and $10.0\text{cm}$ for $d$ in the above equation to calculate $B_{\text{net}}$.

    $\begin{array}{c}{B}_{\text{net}}=-\frac{2\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{\pi \left(10.0\text{cm}\right)}\\ =-\frac{2\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{\pi \left(10.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =-40\times {10}^{-6}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =-40.0\mu \text{T}\end{array}$

Therefore, the magnitude of the net magnetic field is $40.0\mu \text{T}$ and the direction of the net magnetic field is into the page at the mid way between the wires.

(b)

To determine

The magnitude and the direction of the net magnetic field at point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right.

Answer to Problem 19P

The magnitude of the net magnetic field is $5.0\mu \text{T}$ and the direction is out of the page at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right.

Explanation of Solution

Write the expression to obtain the magnetic field due to the wire $1$ at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right.

    $B_1=\frac{{\mu }_{0}i}{2\pi r_1}$

Here, $B_1$ is the magnetic field due to the wire $1$ at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right and $r_1$ is the distance between the wire $1$ and the point where magnetic field is to be determined.

Write the expression to obtain the magnetic field due to the wire $2$ at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right.

    $B_2=\frac{{\mu }_{0}i}{2\pi r_2}$

Here, $B_2$ is the magnetic field due to the wire $2$ at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right and $r_2$ is the distance between the wire $2$ and the point where magnetic field is to be determined.

Write the expression to obtain the net magnetic field at the mid way between the wires.

    $B_{\text{net}}=B_1-B_2$

Here, $B_{\text{net}}$ is the net magnetic field at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right

Substitute $\frac{{\mu }_{0}i}{2\pi r_1}$ for $B_1$ and $\frac{{\mu }_{0}i}{2\pi r_2}$ for $B_2$ in the above equation.

    $\begin{array}{c}{B}_{\text{net}}=\frac{{\mu }_{0}i}{2\pi r_1}-\frac{{\mu }_{0}i}{2\pi r_2}\\ =\frac{{\mu }_{0}i}{2\pi }\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\end{array}$

Further substitute $d$ for $r_1$ and $2d$ for $r_2$ in the above equation.

    $\begin{array}{c}{B}_{\text{net}}=\frac{{\mu }_{0}i}{2\pi }\left(\frac{1}{d}-\frac{1}{2d}\right)\\ =\frac{{\mu }_{0}i}{2\pi }\left(\frac{2-1}{2d}\right)\\ =\frac{{\mu }_{0}i}{4\pi d}\end{array}$

Conclusion:

Substitute $5.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $10.0\text{cm}$ for $d$ in the above equation to calculate $B_{\text{net}}$.

    $\begin{array}{c}{B}_{\text{net}}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{4\pi \left(10.0\text{cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{4\pi \left(10.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =5.0\times {10}^{-6}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =5.0\mu \text{T}\end{array}$

Therefore, the magnitude of the net magnetic field is $5.0\mu \text{T}$ and the direction is out of the page at the point $P_1$ which is $10.0\text{cm}$ to the right of the wire on the right.

(c)

To determine

The magnitude and the direction of the net magnetic field at point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left.

Answer to Problem 19P

The magnitude of the net magnetic field is $1.67\mu \text{T}$ and the direction is out of the page at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left.

Explanation of Solution

Write the expression to obtain the magnetic field due to the wire $1$ at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left.

    $B_1=\frac{{\mu }_{0}i}{2\pi r_1}$

Here, $B_1$ is the magnetic field due to the wire $1$ at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left and $r_1$ is the distance between the wire $1$ and the point where magnetic field is to be determined.

Write the expression to obtain the magnetic field due to the wire $2$ at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left.

    $B_2=\frac{{\mu }_{0}i}{2\pi r_2}$

Here, $B_2$ is the magnetic field due to the wire $2$ at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left and $r_2$ is the distance between the wire $2$ and the point where magnetic field is to be determined.

Write the expression to obtain the net magnetic field at the mid way between the wires.

    $B_{\text{net}}=-B_1+B_2$

Here, $B_{\text{net}}$ is the net magnetic field at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left.

Substitute $\frac{{\mu }_{0}i}{2\pi r_1}$ for $B_1$ and $\frac{{\mu }_{0}i}{2\pi r_2}$ for $B_2$ in the above equation.

    $\begin{array}{c}{B}_{\text{net}}=-\frac{{\mu }_{0}i}{2\pi r_1}+\frac{{\mu }_{0}i}{2\pi r_2}\\ =\frac{{\mu }_{0}i}{2\pi }\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\end{array}$

Further substitute $3d$ for $r_1$ and $2d$ for $r_2$ in the above equation.

    $\begin{array}{c}{B}_{\text{net}}=\frac{{\mu }_{0}i}{2\pi }\left(\frac{1}{2d}-\frac{1}{3d}\right)\\ =\frac{{\mu }_{0}i}{2\pi }\left(\frac{3-2}{6d}\right)\\ =\frac{{\mu }_{0}i}{12\pi d}\end{array}$

Conclusion:

Substitute $5.00\text{A}$ for $i$, $4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}$ for ${\mu }_0$ and $10.0\text{cm}$ for $d$ in the above equation to calculate $B_{\text{net}}$.

    $\begin{array}{c}{B}_{\text{net}}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{12\pi \left(10.0\text{cm}\right)}\\ =\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}/\text{A}\right)\left(5.00\text{A}\right)}{12\pi \left(10.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\\ =1.67\times {10}^{-6}\text{T}\times \frac{{10}^6\mu \text{T}}{1\text{T}}\\ =1.67\mu \text{T}\end{array}$

Therefore, the magnitude of the net magnetic field is $1.67\mu \text{T}$ and the direction is out of the page at the point $P_2$ which is $20.0\text{cm}$ to the left of the wire on the left.