Chapter 30, Problem 35P

(a)

To determine

The distance at which the magnitude of magnetic field is $0.1\text{μT}$.

Answer to Problem 35P

The distance at which the magnitude of magnetic field is $0.1\text{μT}$ is $4.00\text{m}$.

Explanation of Solution

Given info: The magnitude of magnetic field is $1.00\text{μT}$, the distance from the wire is $40.0\text{cm}$ and the current in each wire is $2.00\text{A}$.

The formula to calculate the magnetic field is,

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

Here,

$I$ is the electric current in wire .

$r$ is the distance from the wire.

${\mu }_{\circ }$ is the permeability of free space.

Deduce the formula to calculate the distance from equation (1).

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

Substitute $0.1\text{μT}$ for $B$, $2.00\text{A}$ for $I$ and $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_{\circ }$ in the above equation to find the value of $r$.

$\begin{array}{c}r=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)2.00\text{A}}{2\pi \left(0.1\text{μT}\times \frac{{10}^{-6}\text{T}}{1\text{μT}}\right)}\\ =4.00\text{m}\end{array}$

Thus, the distance at which the magnitude of magnetic field is $0.1\text{μT}$ is $4.00\text{m}$.

Conclusion:

Therefore, the distance at which the magnitude of magnetic field is $0.1\text{μT}$ is $4.00\text{m}$.

(b)

To determine

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

Answer to Problem 35P

The magnetic field $40.0\text{cm}$ away from the middle of the straight cord is $7.50\text{nT}$.

Explanation of Solution

Given info: The magnitude of magnetic field is $1.00\text{μT}$, the distance from the wire is $40.0\text{cm}$ and the current in each wire is $2.00\text{A}$.

Thus, the direction of the magnetic force per unit length on a wire located $0.20\text{cm}$ from the center of the bundle is towards the center as the force is attractive in nature.

The formula to calculate the magnetic field is,

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

Here,

$I$ is the electric current in wire .

$r$ is the distance from the wire.

${\mu }_{\circ }$ is the permeability of free space.

The formula to calculate the net magnetic field is,

$B_{\text{net}}=B_2-B_1$

$B_{\text{net}}=\frac{{\mu }_{\circ }I}{2\pi r_2}-\frac{{\mu }_{\circ }I}{2\pi r_1}$ (1)

The distance is measured from the center. The value of $r_1$ is calculate as,

$\begin{array}{c}{r}_1=40\text{cm+}\frac{1}{2}\left(3\text{mm}\times \frac{{10}^{-1}\text{cm}}{1\text{mm}}\right)\\ =40.15\text{cm}\end{array}$

The value of $r_2$ is calculate as,

$\begin{array}{c}{r}_2=40\text{cm}-\frac{1}{2}\left(3\text{mm}\times \frac{{10}^{-1}\text{cm}}{1\text{mm}}\right)\\ =39.85\text{cm}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_{\circ }$, $2.00\text{A}$ for $I$, $40.15\text{cm}$ for $r_1$ and $39.85\text{cm}$ for $r_2$ in equation (1) to find the value of $B_{\text{net}}$.

$\begin{array}{c}{B}_{\text{net}}=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(39.85\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}-\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(40.15\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =7.50\times {10}^{-9}\text{T}\times \frac{{10}^9\text{nT}}{1\text{T}}\\ =7.50\text{nT}\end{array}$

Thus, the magnetic field $40.0\text{cm}$ away from the middle of the straight cord is $7.50\text{nT}$.

Conclusion:

Therefore, the magnetic field $40.0\text{cm}$ away from the middle of the straight cord is $7.50\text{nT}$.

(c)

To determine

The distance at which the magnetic field is one tenth of $7.50\text{nT}$.

Answer to Problem 35P

The distance at which the magnetic field is one tenth of $7.50\text{nT}$ is $1.26\text{m}$.

Explanation of Solution

Given info: The magnitude of magnetic field is $1.00\text{μT}$, the distance from the wire is $40.0\text{cm}$ and the current in each wire is $2.00\text{A}$.

The formula to calculate the net magnetic field is,

$\begin{array}{c}{B^{\prime }}_{\text{net}}=B_2-B_1\\ =\frac{{\mu }_{\circ }I}{2\pi r_2}-\frac{{\mu }_{\circ }I}{2\pi r_1} \end{array}$ . (1)

The new value of $B_{\text{net}}$ is,

${B^{\prime }}_{\text{net}}=\frac{B_{\text{net}}}{10}$

Substitute $7.50\text{nT}$ for $B_{\text{net}}$ in the above equation to find the value of ${B^{\prime }}_{\text{net}}$.

$\begin{array}{c}{B^{\prime }}_{\text{net}}=\frac{7.50\text{nT}}{10}\\ =0.75\text{nT}\end{array}$

Consider a distance $x$ at which the magnetic field is one tenth. The value of $r_1$ is calculated as,

$\begin{array}{c}{r}_1=x\text{cm+}\frac{1}{2}\left(3\text{mm}\times \frac{{10}^{-1}\text{cm}}{1\text{mm}}\right)\\ =\left(x+0.15\right)\text{cm}\end{array}$

The value of $r_2$ is calculated as,

$\begin{array}{c}{r}_2=x\text{cm-}\frac{1}{2}\left(3\text{mm}\times \frac{{10}^{-1}\text{cm}}{1\text{mm}}\right)\\ =\left(x-0.15\right)\text{cm}\end{array}$

Substitute $4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}$ for ${\mu }_{\circ }$, $2.00\text{A}$ for $I$, $0.75\text{nT}$ for ${B^{\prime }}_{\text{net}}$, $\left(x+0.15\right)\text{cm}$ for $r_1$ and $\left(x-0.15\right)\text{cm}$ for $r_2$ in  equation (1) to find the value of $x$ .

$\begin{array}{c}\left(0.75\text{nT}\right)=\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(x-0.15\right)\text{cm}\right)}-\frac{\left(4\pi \times {10}^{-7}\text{T}\text{.m}/\text{A}\right)\left(2.00\text{A}\right)}{2\pi \left(\left(x+0.15\right)\text{cm}\right)}\\ \left(0.75\text{nT}\times \frac{{10}^{-9}\text{T}}{1\text{nT}}\right)=4\times {10}^{-7}\left(\frac{1}{\left(x-0.15\right)}-\frac{1}{\left(x+0.15\right)}\right)\\ 0.00187=\frac{0.3}{x^2-{0.15}^2}\\ x^2-0.0225=160\end{array}$

Further solve for $x$ .

$\begin{array}{c}x=\left(12.6\text{cm}\times \frac{{10}^{-1}\text{m}}{1\text{cm}}\right)\\ x=1.26\text{m}\end{array}$

Thus, the distance at which the magnetic field is one tenth of $7.50\text{nT}$ is $1.26\text{m}$.

Conclusion:

Therefore, The distance at which the magnetic field is one tenth of $7.50\text{nT}$ is $1.26\text{m}$.

(d)

To determine

The magnetic field the cable create at points outside the cable.

Answer to Problem 35P

The magnetic field that the cable creates at points outside the cable is $0$.

Explanation of Solution

The magnitude of magnetic field is $1.00\text{μT}$, the distance from the wire is $40.0\text{cm}$ and the current in each wire is $2.00\text{A}$.

The formula to calculate the net magnetic field is,

$\begin{array}{c}{B}_{\text{net}}=B_2-B_1\\ =\frac{{\mu }_{\circ }I}{2\pi r_2}-\frac{{\mu }_{\circ }I}{2\pi r_1} \end{array}$

The center wire in a coaxial cable carries a current is $2.00\text{A}$ and the current in the sheath around is $2.00\text{A}$ in the opposite direction. The radius of the center wire is almost equal to that of the sheath around it as the radius of the sheath is negligible. So the value of current and distance is same due to which the value of $B_{\text{net}}$ is zero.

$\begin{array}{c}{B}_{\text{net}}=\frac{{\mu }_{\circ }I}{2\pi r}-\frac{{\mu }_{\circ }I}{2\pi r}\\ =0 \end{array}$

Thus, the magnetic field that the cable creates at points outside the cable is $0$.

Conclusion:

Therefore the magnetic field that the cable creates at points outside the cable is $0$.