Chapter 19, Problem 53P

The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00 μT. (a) At what distance is it 0.100 μT? (b) At one instant, the two conductors in a long household extension cord carry equal 2.00-A currents in opposite directions. The two wires are 3.00 mm apart. Find the magnetic field 40.0 cm away from the middle of the straight cord, in the plane of the two wires. (c) At what distance is it one-tenth as large? (d) The center wire in a coaxial cable carries current 2.00 A in one direction, and the sheath around it carries current 2.00 A in the opposite direction. What magnetic field does the cable create at points outside?

To determine

The distance at which the magnetic field will be $0.100\text{μT}$.

Answer to Problem 53P

The direction of the magnetic field of the wire at the position of the proton is along the negative y-direction.

Explanation of Solution

Given Info: The current in the long straight wire is $2.00\text{A}$. At $40.0\text{cm}$ from the wire the magnetic field is $1.00\text{μΤ}$.

Explanation:

The magnetic field in terms of the current and distance from the wire is,

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

• $I$ is the current
• $r$ is the distance

The distance $r$ will be,

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

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $2.00\text{A}$ for $I$, $1.00\text{μΤ}$ for $B$ to determine the distance $r$,

$\begin{array}{c}r=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.00\text{A}\right)}{2\pi \left(1.00\text{μΤ}\right)}\\ =4.00\text{m}\end{array}$

Conclusion: The distance at which the magnetic field will be $0.100\text{μT}$ is $4.00\text{m}$.

To determine

The magnetic field $40.0\text{cm}$ away from the middle of the straight cord in the plane of the two wires.

Answer to Problem 53P

The net magnetic field at the point which is $40.0\text{cm}$ from the middle of the two wires is $\text{7}\text{.50}\text{nT}$.

Explanation of Solution

Given Info: The two conductors in a household extension cord carry current $2.00\text{A}$ and in opposite direction. The distance between the two wires is $3.00\text{mm}$. The point where magnetic field has to be measured is $40.0\text{cm}$ away from the middle of the straight cord in the plane of the two wires.

Explanation:

The magnetic field in terms of the current and distance from the wire is,

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

• $I$ is the current
• $r$ is the distance

The directions of the currents in the wires are opposite to each other. Hence the net magnetic field due to the two wires will be,

$B_{net}=B_1-B_2$

• $B_1$ is the magnetic field due to the first wire
• $B_2$ is the magnetic field due to the second wire

Using (1) the magnetic field will be,

$B_{net}=\frac{{\mu }_{0}I}{2\pi }\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

• $r_1$ is the distance of the point from the first wire
• $r_2$ is the distance of the point from the second wire

The point is $40.0\text{cm}$ from the middle of the two wires which are $3.00\text{mm}$ apart, hence from the first wire the point will be $40.0\text{cm-1}\text{.50}\text{mm=39}\text{.85}\text{cm}$ far, and from the second wire the point will be $40.0\text{cm+1}\text{.50}\text{mm=40}\text{.15}\text{cm}$ far.

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $2.00\text{A}$ for $I$, $\text{39}\text{.85}\text{cm}$ for $r_1$ and $\text{40}\text{.15}\text{cm}$ for $r_2$ to determine the net magnetic field,

$\begin{array}{c}{B}_{net}=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.00\text{A}\right)}{2\pi }\left(\frac{1}{\text{39}\text{.85}\text{cm}}-\frac{1}{\text{40}\text{.15}\text{cm}}\right)\\ =7.50\times {10}^{-9}\text{T}\\ \text{=7}\text{.50}\text{nT}\end{array}$

Conclusion: The net magnetic field at the point which is $40.0\text{cm}$ from the middle of the two wires is $\text{7}\text{.50}\text{nT}$.

To determine

The distance at which the magnetic field will be one-tenth as large as found in section (b).

Answer to Problem 53P

The distance at which the magnetic field will be one-tenth as large as found in section (b). is $1.26\text{m}$.

Explanation of Solution

Given Info: The two conductors in a household extension cord carry current $2.00\text{A}$ and in opposite direction. The distance between the two wires is $3.00\text{mm}$. The magnetic field to be measured at the point is one tenth of $\text{7}\text{.50}\text{nT}$.

Explanation:

The magnetic field in terms of the current and distance from the wire is,

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

• $I$ is the current
• $r$ is the distance

The directions of the currents in the wires are opposite to each other. Hence the net magnetic field due to the two wires will be,

$B_{net}=B_1-B_2$

• $B_1$ is the magnetic field due to the first wire
• $B_2$ is the magnetic field due to the second wire

Using (1) the magnetic field will be,

$B_{net}=\frac{{\mu }_{0}I}{2\pi }\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

• $r_1$ is the distance of the point from the first wire
• $r_2$ is the distance of the point from the second wire

Consider $r$ is the distance of the point from the centre of the two cords and the distance between the two cords is $2d$. The magnetic field will be,

$B_{net}=\frac{{\mu }_{0}I}{2\pi }\left(\frac{1}{r-d}-\frac{1}{r+d}\right)$

Simplify the above expression,

$B_{net}=\frac{{\mu }_{0}I}{2\pi }\left(\frac{2d}{r^2-d^2}\right)$

The distance from the center $r$ will be,

$r=\sqrt{\frac{{\mu }_{0}Id}{\pi B_{net}}+d^2}$

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}$ for ${\mu }_0$, $2.00\text{A}$ for $I$, $1.50\text{mm}$ for $d$, $7.50\times {10}^{-10}\text{T}$ for $B_{net}$ to determine $r$,

$\begin{array}{c}r=\sqrt{\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.00\text{A}\right)\left(1.50\text{mm}\right)}{\pi \left(7.50\times {10}^{-10}\text{T}\right)}+{\left(1.50\text{mm}\right)}^2}\\ =\sqrt{\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m}\cdot {\text{A}}^{\text{-1}}\right)\left(2.00\text{A}\right)\left(1.50\times {10}^{-3}\text{m}\right)}{\pi \left(7.50\times {10}^{-10}\text{T}\right)}+{\left(1.50\times {10}^{-3}\text{m}\right)}^2}\\ =1.26\text{m}\end{array}$

Conclusion: The distance at which the magnetic field will be one-tenth as large as found in section (b). is $1.26\text{m}$.

To determine

The magnetic field created by the cable at points outside the cable.

Answer to Problem 53P

The magnetic field outside the cable will be zero.

Explanation of Solution

Given Info: The center wire of the coaxial cable carries a current of $2.00\text{A}$. The sheath around the wire also carries $2.00\text{A}$ in opposite direction to the center wire.

Explanation:

Since the center wire and the sheath carries equal and opposite current, for an amperian loop outside the cable, the total enclosed current will be zero. Hence according to the ampere law the magnetic field outside the cable the magnetic field due to cable will be zero.

Conclusion: The magnetic field outside the cable will be zero.