Chapter 22, Problem 45P

(a)

To determine

The magnitude of the magnetic field created by $I_1$ at the location of  $I_2$ .

Answer to Problem 45P

The magnitude of the magnetic field created by $I_1$ at the location $I_2$ is $10\text{μT}$ .

Explanation of Solution

Write the expression of the magnetic field at the origin due to the current $I_1$ .

  $\begin{array}{c}{B}_1=\frac{{\mu }_0{I}_1}{2\pi a}\\ =\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{2I_1}{a}\right)\end{array}$        (I)

Here, $a$ is the distance between the wire and the origin, ${\mu }_0$ is the permeability of free space, and $I_1$ current in the first wire.

Conclusion:

Substitute $1\times {10}^{-7}{\text{N/A}}^2$  for ${\mu }_0/4\pi$, $5.00\text{A}$ for $I_1$,  $10.0\text{cm}$ for $a$ in the above expression (I),

    $\begin{array}{c}{B}_1=\left(\frac{1\times {10}^{-7}{\text{N/A}}^2}{10.0\text{cm}}\right)\left(\frac{2\times 5.00\text{A}}{\raisebox{1ex}{$1\text{m}$}\!\left/ \!\raisebox{-1ex}{$100\text{cm}$}\right.}\right)\\ =1\times {10}^{-5}\text{T}\left(\frac{1\text{μT}}{{10}^{-6}\text{T}}\right)\\ =10\text{μT}\end{array}$                                       

The magnitude of the magnetic field created by $I_1$ at the location $I_2$ is $10\text{μT}$ .

(b)

To determine

The force per unit length

Answer to Problem 45P

The force per unit length exerted by $I_1$ on $I_2$ is $80\text{μN}$ towards the first wire.

Explanation of Solution

Write the expression of the force exerted per unit length by $I_1$ on $I_2$ .

  $F_1=B_1{I}_2$        (I)

Here, $F_1$ is the force per unit length, $B_1$ is the magnetic field due to the current $I_1$ on $I_2$ .

Conclusion:

Substitute $1\times {10}^{-5}\text{T}$  for $B_1$, $8.00\text{A}$ for $I_2$,  in the above expression (I),

    $\begin{array}{c}{F}_1=\left(1\times {10}^{-5}\text{T}\right)\left(8.00\text{A}\right)\\ =8\times {10}^{-5}\text{T}\times \frac{{10}^6\text{μN}}{1\text{N}}\\ =80\text{μN}\end{array}$                                       

The force per unit length exerted by $I_1$ on $I_2$ is $80\text{μN}$ towards the first wire.

(c)

To determine

The magnitude of the magnetic field created by $I_2$ at the location of  $I_1$ .

Answer to Problem 45P

The magnitude of the magnetic field created by $I_2$ at the location $I_1$ is $16\text{μT}$ .

Explanation of Solution

Write the expression of the magnetic field at the origin due to the current $I_1$ .

  $\begin{array}{c}{B}_1=\frac{{\mu }_0{I}_1}{2\pi a}\\ =\left(\frac{{\mu }_0}{4\pi }\right)\left(\frac{2I_1}{a}\right)\end{array}$        (I)

Here, $a$ is the distance between the wire and the origin, ${\mu }_0$ is the permeability of free space, and $I_1$ current in the first wire.

Conclusion:

Substitute $1\times {10}^{-7}\text{N}\text{/}{\text{A}}^2$  for ${\mu }_0/4\pi$, $8.00\text{A}$ for $I_2$,  $10.0\text{cm}$ for $a$ in the above expression (I),

    $\begin{array}{c}{B}_1=\left(\frac{1\times {10}^{-7}{\text{N/A}}^2}{10.0\text{cm}}\right)\left(\frac{2\times 8.00\text{A}}{\raisebox{1ex}{$1\text{m}$}\!\left/ \!\raisebox{-1ex}{$100\text{cm}$}\right.}\right)\\ =1\times {10}^{-5}\text{T}\left(\frac{{10}^6\text{μT}}{1\text{T}}\right)\\ =16\text{μT}\end{array}$                                       

The magnitude of the magnetic field created by $I_2$ at the location $I_1$ is $16\text{μT}$ .

(d)

To determine

The force per unit length

Answer to Problem 45P

The force per unit length exerted by $I_2$ on $I_1$ is $80\text{μN}$ towards the first wire.

Explanation of Solution

Write the expression of the force exerted per unit length by $I_2$ on $I_1$ .

  $F_2=B_1{I}_1$        (I)

Here, $F_2$ is the force per unit length, $B_1$ is the magnetic field due to the current $I_1$ on $I_2$ .

Conclusion:

Substitute $1\times {10}^{-5}\text{T}$  for $B_1$, $5.00\text{A}$ for $I_1$,  in the above expression (I),

    $\begin{array}{c}{F}_1=\left(1.6\times {10}^{-5}\text{T}\right)\left(5.00\text{A}\right)\\ =8\times {10}^{-5}\text{T}\times \frac{{10}^6\text{μN}}{1\text{N}}\\ =80\text{μN}\end{array}$                                       

The force per unit length exerted by $I_2$ on $I_1$ is $80\text{μN}$ towards the first wire.