Chapter 30, Problem 65PQ

(a)

To determine

Find the force per unit length exerted on wire A.

Answer to Problem 65PQ

The force per unit length exerted on wire A is $\left(1.04\times {10}^{-6}\widehat{i}-5.39\times {10}^{-6}\widehat{j}\right)\text{N}/\text{m}$.

Explanation of Solution

The curent flowing in all the wires is in the same direction. Therefore, the wires will be attracted towards each other.

The forces acting on the wire A due to the current in the wire B and wire C are as shown below.

Write the expression for force exerted on wire A due to current in wire C as.

    ${\overrightarrow{F}}_{CA}=\frac{{\mu }_0{I}_A{I}_C}{2\pi r}$                                                                                                  (I)

Here, $I_A$ is current in wire A, $I_C$ is current in wire C, ${\mu }_0$ is the permeability, $r$ is the distance between the wires, ${\overrightarrow{F}}_{CA}$ is force acting on wire A due to wire C.

Write the expression for force exerted on wire A due to current in wire B as.

    ${\overrightarrow{F}}_{BA}=\frac{{\mu }_0{I}_A{I}_B}{2\pi r}$                                                                                                 (II)

Here, $I_B$ is current in wire $B$  and ${\overrightarrow{F}}_{BA}$ is force acting on wire A due to wire B.

The components of the forces ${\overrightarrow{F}}_{CA}$ and ${\overrightarrow{F}}_{BA}$ along the X-axis and Y-axis are given as.

Write the expression for component of the forces along the X-axis as.

  ${\overrightarrow{F}}_x=\left({\overrightarrow{F}}_{CA}\sin \frac{\theta }{2}-{\overrightarrow{F}}_{BA}\sin \frac{\theta }{2}\right)\widehat{i}$                                                                     (III)

Here, ${\overrightarrow{F}}_x$ is the net force in X-axis and $\theta$ is the angle between the two forces.

Write the expression for component of the forces along the Y-axis as.

  ${\overrightarrow{F}}_y=-\left({\overrightarrow{F}}_{CA}\cos \frac{\theta }{2}+{\overrightarrow{F}}_{BA}\cos \frac{\theta }{2}\right)\widehat{j}$                                                                (IV)

Here, ${\overrightarrow{F}}_y$ is the net force in Y-axis.

Write the expression for the net force on the wire A as.

  ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_x+{\overrightarrow{F}}_y$                                                                                                 (V)

Here, ${\overrightarrow{F}}_{net}$ is the net force on the wire A.

The distance between all the three wires is same. Therefore the wires form an equilateral triangle. So, the angle between the two forces is $60°$.

Conclusion:

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}$ for ${\mu }_0$, $1.63\text{ A}$ for $I_A$, $3.26\text{ A}$ for $I_C$ and $0.256\text{ m}$ for $r$ in equation (I).

  $\begin{array}{c}{\overrightarrow{F}}_{CA}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}\right)\left(1.63\text{A}\right)\left(3.26\text{A}\right)}{2\pi \left(0.256\text{m}\right)}\\ =\frac{1.06\times {10}^{-6}}{0.256}\text{N}/\text{m}\\ =4.15\times {10}^{-6}\text{N}/\text{m}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^{\text{2}}}$ for ${\mu }_0$, $1.63\text{ A}$ for $I_A$, $1.63\text{ A}$ for $I_B$ and $0.256\text{ m}$ for $r$ in equation (II)

    $\begin{array}{c}{\overrightarrow{F}}_{BA}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}\right)\left(1.63\text{A}\right)\left(1.63\text{A}\right)}{2\pi \left(0.256\text{m}\right)}\\ =\frac{5.31\times {10}^{-6}}{0.256}\text{N}/\text{m}\\ =2.07\times {10}^{-6}\text{N}/\text{m}\end{array}$

Substitute $4.15\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{CA}$, $2.07\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{BA}$ and $60°$ for $\theta$ in equation (III).

    $\begin{array}{c}{\overrightarrow{F}}_x=\left(\left(4.15\times {10}^{-6}\text{N}/\text{m}\right)\sin 30°-\left(2.07\times {10}^{-6}\text{N}/\text{m}\right)\sin 30°\right)\widehat{i}\\ =\left(\left(2.075\times {10}^{-6}\text{N}/\text{m}\right)-\left(1.035\times {10}^{-6}\text{N}/\text{m}\right)\right)\widehat{i}\\ =\left(1.04\times {10}^{-6}\text{N}/\text{m}\right)\widehat{i}\end{array}$

Substitute $4.15\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{CA}$, $2.07\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{BA}$ and $60°$ for $\theta$  in equation (IV).

  $\begin{array}{c}{\overrightarrow{F}}_y=-\left(\left(4.15\times {10}^{-6}\text{N}/\text{m}\right)\cos 30°+\left(2.07\times {10}^{-6}\text{N}/\text{m}\right)\cos 30°\right)\widehat{j}\\ =-\left(\left(3.594\times {10}^{-6}\text{N}/\text{m}\right)+\left(1.792\times {10}^{-6}\text{N}/\text{m}\right)\right)\widehat{j}\\ =-\left(5.386\times {10}^{-6}\text{N}/\text{m}\right)\widehat{j}\\ \approx -\left(5.39\times {10}^{-6}\text{N}/\text{m}\right)\widehat{j}\end{array}$

Substitute $1.04\times {10}^{-6}\widehat{i}\text{N}/\text{m}$ for ${\overrightarrow{F}}_x$ and $-5.39\times {10}^{-6}\widehat{j}\text{N}/\text{m}$ for ${\overrightarrow{F}}_y$ in equation (V).

  ${\overrightarrow{F}}_{net}=\left(1.04\times {10}^{-6}\widehat{i}-5.39\times {10}^{-6}\widehat{j}\right)\text{N}/\text{m}$

Thus, the force per unit length exerted on wire A is $\left(1.04\times {10}^{-6}\widehat{i}-5.39\times {10}^{-6}\widehat{j}\right)\text{N}/\text{m}$.

(b)

To determine

Find the force per unit length exerted on wire B.

Answer to Problem 65PQ

The force per unit length exerted on wire B is $\left(5.19\times {10}^{-6}\widehat{i}+1.80\times {10}^{-6}\widehat{j}\right)\text{N}/\text{m}$.

Explanation of Solution

The forces acting on the wire A due to the current in the wire B and wire C are as shown below.

Write the expression for force exerted on wire B due to current in wire A as.

  ${\overrightarrow{F}}_{AB}=\frac{{\mu }_0{I}_A{I}_B}{2\pi r}$                                                                                              (VI)

Here, ${\overrightarrow{F}}_{AB}$ is the force acting on wire B due to wire A.

Write the expression for force exerted on wire B due to current in wire Cas.

    ${\overrightarrow{F}}_{CB}=\frac{{\mu }_0{I}_C{I}_B}{2\pi r}$                                                                                              (VII)

Here, ${\overrightarrow{F}}_{CB}$ is the force acting on wire B due to wire C.

The components of the forces ${\overrightarrow{F}}_{AB}$ and ${\overrightarrow{F}}_{CB}$ along the X-axis and Y-axis are given as.

Write the expression for component of the forces along the X-axis as.

  ${\overrightarrow{F}}_x=\left({\overrightarrow{F}}_{CB}+{\overrightarrow{F}}_{AB}\cos \theta \right)\widehat{i}$                                                                           (VIII)

Here, ${\overrightarrow{F}}_x$ is the net force in X-axis and $\theta$ is the angle between the two forces.

Write the expression for component of the forces along the Y-axis as.

  ${\overrightarrow{F}}_y=\left({\overrightarrow{F}}_{AB}\sin \theta \right)\widehat{j}$                                                                                       (IX)

Here, ${\overrightarrow{F}}_y$ is the net force in Y-axis.

Write the expression for the net force on the wire B as.

  ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_x+{\overrightarrow{F}}_y$                                                                                               (X)

Here, ${\overrightarrow{F}}_{net}$ is the net force on the wire B.

The wires form an equilateral triangle. So, the angle between the two forces is $60°$.

Conclusion:

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}$ for ${\mu }_0$, $1.63\text{ A}$ for $I_A$, $\text{1}\text{.63 A}$ for $I_B$ and $0.256\text{ m}$ for $r$ in equation (VI).

  $\begin{array}{c}{\overrightarrow{F}}_{AB}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}\right)\left(1.63\text{ A}\right)\left(1.63\text{ A}\right)}{2\pi \left(0.256\text{ m}\right)}\\ =\frac{5.314\times {10}^{-7}}{0.256}\text{N}/\text{m}\\ =2.07\times {10}^{-6}\text{N}/\text{m}\end{array}$

Substitute $4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}$ for ${\mu }_0$, $\text{3}\text{.26 A}$ for $I_C$, $\text{1}\text{.63 A}$ for $I_B$ and $0.256\text{ m}$ for $r$ in equation (VII).

  $\begin{array}{c}{\overrightarrow{F}}_{CB}=\frac{\left(4\pi \times {10}^{-7}\frac{\text{N}}{{\text{A}}^2}\right)\left(\text{3}\text{.26 A}\right)\left(1.63\text{ A}\right)}{2\pi \left(0.256\text{ m}\right)}\\ =\frac{1.06\times {10}^{-6}}{0.256}\text{N}/\text{m}\\ =4.15\times {10}^{-6}\text{N}/\text{m}\end{array}$

Substitute $4.15\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{CB}$, $2.07\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{AB}$ and $60°$ for $\theta$ in equation (VIII).

  $\begin{array}{c}{\overrightarrow{F}}_x=\left(\left(4.15\times {10}^{-6}\text{N}/\text{m}\right)+\left(2.07\times {10}^{-6}\text{N}/\text{m}\right)\cos 60°\right)\widehat{i}\\ =\left(5.19\times {10}^{-6}\text{N}/\text{m}\right)\widehat{i}\end{array}$

Substitute $2.07\times {10}^{-6}\text{N}/\text{m}$ for ${\overrightarrow{F}}_{AB}$ and $60°$ for $\theta$ in equation (IX).

  $\begin{array}{c}{\overrightarrow{F}}_y=\left(\left(2.07\times {10}^{-6}\text{N}/\text{m}\right)\sin 60°\right)\widehat{j}\\ =\left(1.80\times {10}^{-6}\text{N}/\text{m}\right)\widehat{j}\end{array}$

Substitute $5.19\times {10}^{-6}\widehat{i}\text{N}/\text{m}$ for ${\overrightarrow{F}}_x$ and $1.80\times {10}^{-6}\widehat{j}\text{N}/\text{m}$ for ${\overrightarrow{F}}_y$ in equation (X).

  ${\overrightarrow{F}}_{net}=\left(5.19\times {10}^{-6}\widehat{i}+1.80\times {10}^{-6}\widehat{j}\right)\text{N}/\text{m}$

Thus, the force per unit length exerted on wire B is $\left(5.19\times {10}^{-6}\widehat{i}+1.80\times {10}^{-6}\widehat{j}\right)\text{N}/\text{m}$.