Chapter 22, Problem 20P

In Figure P22.20, the cube is 40.0 cm on each edge. Four straight segments of wire—ab, bc, cd, and da—form a closed loop that carries a current I = 5.00 A in the direction shown. A uniform magnetic field of magnitude B = 0.020 0 T is in the positive y direction. Determine the magnetic force vector on (a) ab, (b) bc, (c) cd, and (d) da. (e) Explain how you could find the force exerted on the fourth of these segments from the forces on the other three, without further calculation involving the magnetic field.

To determine

The magnetic force vector on $ab$ .

Answer to Problem 20P

The magnetic force vector on $ab$ is zero.

Explanation of Solution

Write the expression for the magnetic field in a wire,

  $F=BIL\sin \theta$        (I)

Here, $F$ is the centripetal force, $L$ is the length of the particle, $I$ is the current of the particle, $\theta$ is the angle between the current and the field, and  $B$ is the magnetic field.

Conclusion:

Substitute $0.020\text{T}$ for $B$ , $5.00\text{A}$ for $I$ , $180°$  for $\theta$ and $40.0\text{cm}$ for $l$ in the above equation to find the value of $F$ .

    $\begin{array}{c}F=\left(0.020\text{T}\right)\left(5.00\text{A}\right)\left(40.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 180°\\ =0\end{array}$

The magnetic force vector on $ab$ is zero.

To determine

The magnetic force vector on $bc$ .

Answer to Problem 20P

The magnetic force vector on $bc$ is $0.04\widehat{\text{i}}\text{N}$.

Explanation of Solution

Write the expression for the magnetic field in a wire,

  $F=BIL\sin \theta$        (I)

Here, $F$ is the centripetal force, $L$ is the length of the particle, $I$ is the current of the particle, $\theta$ is the angle between the current and the field, and  $B$ is the magnetic field

Conclusion:

Substitute $0.020\text{T}$ for $B$ , $5.00\text{A}$ for $I$ , $90°$  for $\theta$ and $40.0\text{cm}$ for $l$ in the above equation to find the value of $F$ .

    $\begin{array}{c}F=\left(0.020\text{T}\right)\left(5.00\text{A}\right)\left(40.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 90°\\ =0.04\text{N}\end{array}$

By Flemings right hand rule, the thumb points towards the $x$ -axis , thus the direction of the fore is along $x$-direction.

The magnetic force vector on $bc$ is $0.04\widehat{\text{i}}\text{N}$.

To determine

The magnetic force vector on $cd$ .

Answer to Problem 20P

The magnetic force vector on $cd$ is $0.04\widehat{\text{k}}\text{N}$.

Explanation of Solution

Write the expression for the magnetic field in a wire,

  $F=BIL\sin \theta$        (I)

Here, $F$ is the centripetal force, $L$ is the length of the particle, $I$ is the current of the particle, $\theta$ is the angle between the current and the field, and  $B$ is the magnetic field

Write the Pythagoras equation to obtain the value for $l$ ,

  $\begin{array}{c}l=\sqrt{a^2+a^2}\\ =\sqrt{2a^2}\end{array}$        (II)

Conclusion:

Substitute $40.0\text{cm}$ for $a$ ,in the expression (II),

    $\begin{array}{c}l=\sqrt{2\times 40.0{\text{cm}}^{\text{2}}}\\ =57.0\text{cm}\end{array}$

Substitute $0.020\text{T}$ for $B$ , $5.00\text{A}$ for $I$ , $45°$  for $\theta$ and $57.0\text{cm}$ for $l$ in the above equation (I) to find the value of $F$ .

    $\begin{array}{c}F=\left(0.020\text{T}\right)\left(5.00\text{A}\right)\left(57.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 45°\\ =0.04\text{N}\end{array}$

By Flemings right hand rule, the thumb points towards the $z$ -axis , thus the direction of the fore is along $z$-direction.

The magnetic force vector on $bc$ is $0.04\widehat{\text{i}}\text{N}$.

To determine

The magnetic force vector on $da$ .

Answer to Problem 20P

The magnetic force vector on $da$ is $0.057\left(\frac{1}{\sqrt{2}}\left(\widehat{i}+\widehat{k}\right)\right)\text{N}$.

Explanation of Solution

Write the expression for the magnetic field in a wire,

  $F=BIL\sin \theta$        (I)

Here, $F$ is the centripetal force, $L$ is the length of the particle, $I$ is the current of the particle, $\theta$ is the angle between the current and the field, and  $B$ is the magnetic field

Write the Pythagoras equation to obtain the value for $l$ ,

  $\begin{array}{c}l=\sqrt{a^2+a^2}\\ =\sqrt{2a^2}\end{array}$        (II)

Conclusion:

Substitute $40.0\text{cm}$ for $a$ ,in the expression (II),

    $\begin{array}{c}l=\sqrt{2\times 40.0{\text{cm}}^{\text{2}}}\\ =57.0\text{cm}\end{array}$

Substitute $0.020\text{T}$ for $B$ , $5.00\text{A}$ for $I$ , $9000°$  for $\theta$ and $57.0\text{cm}$ for $l$ in the above equation (I) to find the value of $F$ .

    $\begin{array}{c}F=\left(0.020\text{T}\right)\left(5.00\text{A}\right)\left(57.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 90°\\ =0.057\text{N}\end{array}$

By Flemings right hand rule, the thumb points the direction $d$ .

The direction of the fore is

    $\begin{array}{c}d=\cos 45°\widehat{i}+\cos 45°\widehat{k}\\ =\left(\frac{1}{\sqrt{2}}\left(\widehat{i}+\widehat{k}\right)\right)\end{array}$

The magnetic force vector on $da$ is $0.057\left(\frac{1}{\sqrt{2}}\left(\widehat{i}+\widehat{k}\right)\right)\text{N}$.

To determine

The force exerted on the forth segment from the forces on the other three.

Answer to Problem 20P

The force exerted on the forth segment from the forces on the other three can be determined by parallelogram law of vectors.

Explanation of Solution

 Write the parallelogram law of vector equation.

  ${\overrightarrow{F}}_4={\overrightarrow{F}}_3+{\overrightarrow{F}}_2+{\overrightarrow{F}}_1$ 

Here, ${\overrightarrow{F}}_4$ is the force for the resultant, ${\overrightarrow{F}}_3$ is the force for the third arm, ${\overrightarrow{F}}_2$ is the first arm of parallelogram, and ${\overrightarrow{F}}_1$ is the second arm of the parallelogram.

Conclusion:

The force exerted on the forth segment from the forces on the other three can be determined by parallelogram law of vectors.