Chapter 22, Problem 51P

(a)

To determine

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

Answer to Problem 51P

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

Explanation of Solution

Write the expression for the magnetic field on a wire.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance of the wire.

Re-write the expression (I) in terms of $r$.

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

Conclusion:

Substitute $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$,  $0.1\text{μT}$ for $B$ , and $0.200\text{A}$ for $I$, in the expression (II),

    $\begin{array}{c}r=\frac{\left(4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}\right)0.200\text{A}}{2\pi \left(0.1\text{μT}\times \frac{{10}^{-6}\text{T}}{1\text{μT}}\right)}\\ =4.00\text{m}\end{array}$

The distance at which the magnitude of the 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 51P

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

Explanation of Solution

Write the expression for the magnetic field on a wire.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance of the wire.

Write the formula to calculate the net magnetic field,

  $B_{net}=B_2-B_1$        (II)

Conclusion:

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

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

The value of $r_1$ 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 $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ , $40.15\text{cm}$ for $r_1$,  $39.85\text{cm}$ for $r_2$ and $2.00\text{A}$ for $I$ in the expression (II) to determine the value for the total magnetic field.

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

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 51P

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

Explanation of Solution

Write the expression for the magnetic field on a wire.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance of the wire.

Write the formula to calculate the net magnetic field,

  ${B^{\prime }}_{net}=B_2-B_1$        (II)

Thus the new value for the $B_{net}$ is ,

  ${B^{\prime }}_{net}=\frac{B_{net}}{10}$        (III)

Conclusion:

Substitute $7.50\text{nT}$ for $B_{net}$ in the expression (III),

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

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 for $r_2$ is calculates 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 $4\pi \times {10}^{-7}\text{T}\cdot \text{m/A}$ for ${\mu }_0$ , $\left(x+0.15\right)\text{cm}$ for $r_1$,  $\left(x-0.15\right)\text{cm}$ for $r_2$, $0.75\text{nT}$ for ${B^{\prime }}_{net}$ and $2.00\text{A}$ for $I$ in the expression (II) to determine the value for $x$ .

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

On further solving for $x$ ,

  $\begin{array}{l}x=\sqrt{160-0.225}\\ x=1.26\text{cm}\times \frac{{10}^{-1}\text{m}}{1\text{cm}}\\ x=1.26\text{m}\end{array}$

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 point outside the cable.

Answer to Problem 51P

The magnetic field the cable create at point outside the cable is $0$.

Explanation of Solution

Write the expression for the magnetic field on a wire.

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

Here, $B$ is the magnetic field, ${\mu }_0$ is the permeability of free space, $I$ is the current, and $r$ is the distance of the wire.

Write the formula to calculate the net magnetic field,

  $B_{net}=B_2-B_1$        (II)

Conclusion:

The center wire in a coaxial cable carriers a current of $2.00\text{A}$ an the current in the sheath around is $2.00\text{A}$ in the opposite direction. The radius of the wire in the center is almost equal to the sheath around it and the radius of the sheath is neglected. Thus the value of the current and the distance is the same due to which the value of the total magnetic field is zero.

    $\begin{array}{c}{B}_{net}=B_2-B_1\\ =\frac{{\mu }_{0}I}{2\pi r}-\frac{{\mu }_{0}I}{2\pi r}\\ =0\end{array}$

Thus, the magnetic field the cable create at point outside the cable is $0$.