Chapter 29, Problem 44P

(a)

To determine

The magnetic force vector on $ab$.

Answer to Problem 44P

The magnetic force vector on $ab$ is $0\text{N}$.

Explanation of Solution

Write the equation for the magnetic force vector on the conductor.

    ${\overrightarrow{F}}_{\text{B}}=I\overrightarrow{L}\times \overrightarrow{B}$                                                                                                              (I)

Here, ${\overrightarrow{F}}_{\text{B}}$ is the magnetic force vector, $I$ is current, $\overrightarrow{L}$ is the length vector of the conductor and $\overrightarrow{B}$ is the magnetic field vector.

Conclusion:

Substitute $5.00\text{A}$ for $I$, $-0.400\widehat{\text{j}}\text{m}$ for $\overrightarrow{L}$ and $0.0200\widehat{\text{j}}\text{T}$ for $\overrightarrow{B}$ in the equation (I) to calculate the magnetic force vector.

    $\begin{array}{c}{\overrightarrow{F}}_{\text{B}}=\left[\left(5.00\text{A}\right)\left(-0.400\widehat{\text{j}}\text{m}\times 0.0200\widehat{\text{j}}\text{T}\right)\right]\\ \text{=}\left(5.00\text{A}\right)\left[\left(-0.400\text{m}\right)\left(0.0200\text{T}\right)\left(\widehat{\text{j}}\times \widehat{\text{j}}\right)\right]\\ =0\text{N}\end{array}$

Therefore, the magnetic force vector on $ab$ is $0\text{N}$.

(b)

To determine

The magnetic vector force on $bc$.

Answer to Problem 44P

The magnetic force vector on $bc$ is $-0.040\widehat{\text{i}}\text{N}$.

Explanation of Solution

Conclusion:

Substitute $5.00\text{A}$ for $I$, $0.400\widehat{\text{k}}\text{m}$ for $\overrightarrow{L}$ and $0.0200\widehat{\text{j}}\text{T}$ for $\overrightarrow{B}$ in the equation (I) to calculate the magnetic force vector.

    $\begin{array}{c}{\overrightarrow{F}}_{\text{B}}=\left[\left(5.00\text{A}\right)\left(0.400\widehat{\text{k}}\text{m}\times 0.0200\widehat{\text{j}}\text{T}\right)\right]\\ \text{=}\left(5.00\text{A}\right)\left[\left(0.400\text{m}\right)\left(0.0200\text{T}\right)\left(\widehat{\text{k}}\times \widehat{\text{j}}\right)\right]\\ =\left(5.00\right)\left[-0.008\widehat{\text{i}}\right]\\ \text{=}-0.040\widehat{\text{i}}\text{N}\end{array}$

Therefore, the magnetic vector force on $bc$ $-0.040\widehat{\text{i}}\text{N}$.

(c)

To determine

The magnetic force vector on $cd$.

Answer to Problem 44P

The magnetic force vector on $cd$ is $-0.040\widehat{\text{k}}\text{N}$.

Explanation of Solution

Conclusion:

Substitute $5.00\text{A}$ for $I$, $\left(-0.400\widehat{\text{i}}\text{m}+0.400\widehat{\text{j}}\text{m}\right)$ for $\overrightarrow{L}$ and $0.0200\widehat{\text{j}}\text{T}$ for $\overrightarrow{B}$ in the equation (I) to calculate the magnetic force vector.

    $\begin{array}{c}{\overrightarrow{F}}_{\text{B}}=\left[\left(5.00\text{A}\right)\left[\left(-0.400\widehat{\text{i}}\text{m+0}\text{.4}00\widehat{\text{j}}\text{m}\right)\times 0.0200\widehat{\text{j}}\text{T}\right]\right]\\ \text{=}\left(5.00\text{A}\right)\left[\left(-0.400\text{m}\right)\left(0.0200\text{T}\right)\left(\widehat{\text{i}}\times \widehat{\text{j}}\right)+\left(0.400\text{m}\right)\left(0.0200\text{T}\right)\left(\widehat{\text{j}}\times \widehat{\text{j}}\right)\right]\\ \text{=}\left(5.00\right)\left[-0.008\widehat{\text{k}}+\text{0}\right]\text{N}\\ =-0.040\widehat{\text{k}}\text{N}\end{array}$

Therefore, the magnetic force vector on $cd$ is $-0.040\widehat{\text{k}}\text{N}$.

(d)

To determine

The magnetic force vector on $da$.

Answer to Problem 44P

The magnetic force vector on $da$ is $\left(0.040\widehat{\text{k}}+0.040\widehat{\text{i}}\right)\text{N}$.

Explanation of Solution

Conclusion:

Substitute $5.00\text{A}$ for $I$, $\left(0.400\widehat{\text{i}}\text{m}-\text{0}\text{.4}00\widehat{\text{k}}\text{m}\right)$ for $\overrightarrow{L}$ and $0.0200\widehat{\text{j}}\text{T}$ for $\overrightarrow{B}$ in the equation (I) to calculate the magnetic force vector.

    $\begin{array}{c}{\overrightarrow{F}}_{\text{B}}=\left[\left(5.00\text{A}\right)\left[\left(0.400\widehat{\text{i}}\text{m}-0.400\widehat{\text{k}}\text{m}\right)\times 0.0200\widehat{\text{j}}\text{T}\right]\right]\\ \text{=}\left(5.00\text{A}\right)\left[\left(0.400\text{m}\right)\left(0.0200\text{T}\right)\left(\widehat{\text{i}}\times \widehat{\text{j}}\right)+\left(0.400\text{m}\right)\left(0.0200\text{T}\right)\left(\widehat{\text{k}}\times \widehat{\text{j}}\right)\right]\\ \text{=}\left(5.00\right)\left[0.008\widehat{\text{k}}-0.008\widehat{\text{i}}\right]\text{N}\\ =\left(0.040\widehat{\text{k}}+0.040\widehat{\text{i}}\right)\text{N}\end{array}$

Therefore, the magnetic force vector on $da$ is $\left(0.040\widehat{\text{k}}+0.040\widehat{\text{i}}\right)\text{N}$.

(e)

To determine

The method to find the force exerted on the fourth segment due to the forces on the other three segments.

Answer to Problem 44P

The force exerted on the fourth segment due to the forces on the other three segments is the negative of resultant of the sum of the forces on other three segments.

Explanation of Solution

The sum of the force on all the four segments is equal to zero. Hence, the force on the fourth vector will be equal to the negative of resultant of the sum of the forces on other three segments.

Conclusion:

Therefore, the force exerted on the fourth segment due to the forces on the other three segments is the negative of resultant of the sum of the forces on other three segments.