Chapter 8, Problem 9P

(a)

To determine

Calculate the force per unit length on $L_2$ due to $L_1$. Also, state whether the force is attractive or repulsive.

Answer to Problem 9P

The force per unit length on $L_2$ due to $L_1$ is $F_{21}=1a_x\text{mN}/\text{m}$ and the force is repulsive.

Explanation of Solution

Calculation:

Write the general expression to calculate the force.

$F=IL\times B$

or

$\frac{F}{L}=I_1{a}_l\times B_2=\frac{{\mu }_{\text{o}}{I}_1{I}_2{a}_l\times a_{\varphi }}{2\pi \rho }\left\{\because B_2=\frac{{\mu }_{\text{o}}{I}_2}{2\pi \rho }{a}_{\varphi }\right\}$        (1)

Here,

${\mu }_{\text{o}}$ is the permeability of free space,

$I_1$ and $I_2$ are the currents, and

$B_2$ is the magnetic flux density.

To find the distance $\rho$ at points (0, 4) and (0, 0).

$\begin{array}{c}\rho =\sqrt{{\left(0-0\right)}^2+{\left(0-4\right)}^2}\\ =4\end{array}$

Using equation (1), the force per unit length on $L_2$ due to $L_1$ is calculated as follows.

$\begin{array}{c}{F}_{21}=\frac{\left(4\pi \times {10}^{-7}\right)\left(-100\right)\left(200\right)a_z\times \left(-a_y\right)}{2\pi \left(4\right)}\left\{\because \rho =4,{\mu }_{\text{o}}=4\pi \times {10}^{-7}\right\}\\ =1\times {10}^{-3}{a}_x\left\{\because a_z\times \left(-a_y\right)=a_x\right\}\\ F_{21}=1a_x\text{mN}/\text{m}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

The force per unit length on $L_2$ due to $L_1$ is repulsive because the current 200 A and –100 A flows in opposite direction.

Conclusion:

Thus, the force per unit length on $L_2$ due to $L_1$ is $F_{21}=1a_x\text{mN}/\text{m}$ and the force is repulsive.

(b)

To determine

Calculate the force per unit length on $L_1$ due to $L_2$. Also, state whether the force is attractive or repulsive.

Answer to Problem 9P

The force per unit length on $L_1$ due to $L_2$ is $F_{12}=-1a_x\text{mN}/\text{m}$ and the force is repulsive.

Explanation of Solution

Calculation:

Refer to Part (a),

The force per unit length on $L_2$ due to $L_1$ is $F_{21}=1a_x\text{mN}/\text{m}$.

The force per unit length on $L_1$ due to $L_2$ is calculated as follows.

$F_{12}=-F_{21}$

Substitute $1a_x\text{mN}/\text{m}$ for $F_{21}$ in above equation.

$F_{12}=-1a_x\text{mN}/\text{m}$

The force per unit length on $L_1$ due to $L_2$ is repulsive because the current 200 A and –100 A flows in opposite direction.

Conclusion:

Thus, the force per unit length on $L_1$ due to $L_2$ is $F_{12}=-1a_x\text{mN}/\text{m}$ and the force is repulsive.

(c)

To determine

Calculate the force per unit length on $L_3$ due to $L_1$. Also, state whether the force is attractive or repulsive.

Answer to Problem 9P

The force per unit length on $L_3$ due to $L_1$ is $F_{31}=\left(-0.96a_x+0.72a_y\right)\text{mN}/\text{m}$ and the force is repulsive.

Explanation of Solution

Calculation:

The unit vector $a_{\varphi }$ is determined as follows,

$\begin{array}{c}{a}_{\varphi }=\frac{\left(0,0\right)-\left(3,4\right)}{\sqrt{{\left(0-3\right)}^2+{\left(0-4\right)}^2}}\\ =\frac{-3a_x-4a_y}{\sqrt{9+16}}\\ =\frac{-3a_x-4a_y}{5}\\ =\frac{-3a_x}{5}-\frac{4a_y}{5}\end{array}$

In equation (1), $a_l\times a_{\varphi }$ is determined as follows.

$\begin{array}{c}{a}_l\times a_{\varphi }=a_z\times \left(-\frac{3a_x}{5}-\frac{4a_y}{5}\right)\\ =-\frac{3}{5}{a}_y+\frac{4}{5}{a}_x\end{array}$

To find the distance $\rho$ at points (3, 4) and (0, 0).

$\begin{array}{c}\rho =\sqrt{{\left(0-3\right)}^2+{\left(0-4\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

Using equation (1), the force per unit length on $L_3$ due to $L_1$ is calculated as follows.

$\begin{array}{c}{F}_{31}=\frac{\left(4\pi \times {10}^{-7}\right)\left(-100\right)\left(300\right)\left(-\frac{3}{5}{a}_y+\frac{4}{5}{a}_x\right)}{2\pi \left(5\right)}\left\{\begin{array}{l}\because a_l\times a_{\varphi }=-\frac{3}{5}{a}_y+\frac{4}{5}{a}_x,\\ \rho =5,{\mu }_{\text{o}}=4\pi \times {10}^{-7}\end{array}\right\}\\ =-1.2\times {10}^{-3}\left(\frac{4}{5}{a}_x-\frac{3}{5}{a}_y\right)\\ F_{31}=\left(-0.96a_x+0.72a_y\right)\text{mN}/\text{m}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

The force per unit length on $L_3$ due to $L_1$ is repulsive because the current 300 A and –100 A flows in opposite direction.

Conclusion:

Thus, the force per unit length on $L_3$ due to $L_1$ is $F_{31}=\left(-0.96a_x+0.72a_y\right)\text{mN}/\text{m}$ and the force is repulsive.

(d)

To determine

Calculate the force per unit length on $L_3$ due to $L_1$ and $L_2$. Also, state whether the force is attractive or repulsive.

Answer to Problem 9P

The force per unit length on $L_3$ due to $L_1$ and $L_2$ is $F_3=-4.96a_x+0.72a_y\text{mN}/\text{m}$. The force per unit length on $L_3$ due to $L_1$ is repulsive and the force per unit length on $L_3$ due to $L_2$ is attractive.

Explanation of Solution

Calculation:

To find the force per unit length on $L_3$ due to $L_2$:

To find the distance $\rho$ at points (3, 4) and (0, 4).

$\begin{array}{c}\rho =\sqrt{{\left(0-3\right)}^2+{\left(4-4\right)}^2}\\ =\sqrt{9+0}\\ =3\end{array}$

Using equation (1), the force per unit length on $L_3$ due to $L_2$ is calculated as follows.

$\begin{array}{c}{F}_{32}=\frac{\left(4\pi \times {10}^{-7}\right)\left(200\right)\left(300\right)a_z\times a_y}{2\pi \left(3\right)}\left\{\because \rho =3,{\mu }_{\text{o}}=4\pi \times {10}^{-7}\right\}\\ =-4\times {10}^{-3}{a}_x\left\{\because a_z\times a_y=-a_x\right\}\\ F_{32}=-4a_x\text{mN}/\text{m}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

The force per unit length on $L_3$ due to $L_1$ and $L_2$ is,

$F_3=F_{31}+F_{32}$

Substitute $\left(-0.96a_x+0.72a_y\right)\text{mN}/\text{m}$ for $F_{31}$ and $-4a_x\text{mN}/\text{m}$ for $F_{32}$ in above equation.

$\begin{array}{c}{F}_3=\left(-0.96a_x+0.72a_y\right)\text{mN}/\text{m}-4a_x\text{mN}/\text{m}\\ =-4.96a_x+0.72a_y\text{mN}/\text{m}\end{array}$

The force per unit length on $L_3$ due to $L_1$ is repulsive because the current 300 A and –100 A flows in opposite direction. The force per unit length on $L_3$ due to $L_2$ is attractive because the current 300 A and 200 A flows in same direction

Conclusion:

Thus, the force per unit length on $L_3$ due to $L_1$ and $L_2$ is $F_3=-4.96a_x+0.72a_y\text{mN}/\text{m}$. The force per unit length on $L_3$ due to $L_1$ is repulsive and the force per unit length on $L_3$ due to $L_2$ is attractive.