Chapter 28, Problem 28P

In Figure P28.28, 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. (c) 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.

Figure P28.28

To determine

The magnetic force vector on $ab$.

Answer to Problem 28P

The magnetic force vector on $ab$ is 0.

Explanation of Solution

Given info: The length of each edge of the cube is $40.0\text{cm}$, the electric current is $5.00\text{A}$ and the magnitude of magnetic field is $0.020\text{T}$.

The formula for the magnetic force is,

$F=BIl\sin \theta$

Here,

$B$ is the magnitude of magnetic field.

$I$ is the intensity of electric current.

$l$ is the length of the cube side $ab$.

$\theta$ is the angle between length and magnetic field.

As the current flows down and the magnetic from point $b$ is upwards so the angle $\theta$ is $180°$.

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 magnitude of the force is zero because an equal and opposite force cancels it due to which the magnetic force of $ab$ have no direction.

Conclusion:

Therefore, magnetic force vector on $ab$ is 0.

To determine

The magnetic force vector on $bc$.

Answer to Problem 28P

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

Explanation of Solution

Given info: The length of each edge of the cube is $40.0\text{cm}$, the electric current is $5.00\text{A}$ and the magnitude of magnetic field is $0.020\text{T}$

The formula for the magnetic force is,

$F=BIl\sin \theta$

Here,

$B$ is the magnitude of magnetic field.

$I$ is the intensity of electric current.

$l$ is the length of the cube side $bc$.

$\theta$ is the angle between length and magnetic field.

As the current flows down and the magnetic from point $c$ is perpendicular so the angle $\theta$ is $90°$.

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 using the Flemings right hand rule, the thumb points towards the x axis direction. Thus, the direction of the magnetic force is in $x$-direction.

Conclusion:

Therefore, 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 28P

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

Explanation of Solution

Given info: The length of each edge of the cube is $40.0\text{cm}$, the electric current is $5.00\text{A}$ and the magnitude of magnetic field is $0.020\text{T}$

The formula for the magnetic force is,

$F=BIl\sin \theta$

Here,

$B$ is the magnitude of magnetic field.

$I$ is the intensity of electric current.

$l$ is the length of the cube side $cd$.

$\theta$ is the angle between length and magnetic field.

Use Pythagoras theorem to find $l$,

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

Substitute $40.0\text{cm}$ for $a$ in the above equation to find the value of $l$.

$\begin{array}{c}l=\sqrt{2\times 40.0{\text{cm}}^2}\\ =57\text{cm}\end{array}$

As the current flows at an angle and the magnetic from point b is perpendicular so the angle $\theta$ is $45°$.

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 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 using the Flemings right hand rule, the thumb points towards the z axis direction. Thus, the direction of the magnetic force is in z direction.

Conclusion:

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

To determine

The magnetic force vector on $da$.

Answer to Problem 28P

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

Explanation of Solution

Given info: The length of each edge of the cube is $40.0\text{cm}$, the electric current is $5.00\text{A}$ and the magnitude of magnetic field is $0.020\text{T}$

The formula for the magnetic force is,

$F=BIl\sin \theta$

Here,

$B$ is the magnitude of magnetic field.

$I$ is the intensity of electric current.

$l$ is the length of the cube side $da$.

$\theta$ is the angle between length and magnetic field.

Use Pythagoras theorem to find $l$,

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

Substitute $40.0\text{cm}$ for $a$ in the above equation to find the value of $l$.

$\begin{array}{c}l=\sqrt{2\times 40.0{\text{cm}}^2}\\ =57\text{cm}\end{array}$

As the current flows vertically and the magnetic from point b is perpendicular so the angle $\theta$ is $90°$.

Substitute $0.020\text{T}$ for $B$, $5.00\text{A}$ for $I$, $90°$ for $\theta$ and $57.0\text{cm}$ for $l$ in the above equation to find the value of $F$.

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

By using the Flemings right hand rule, the thumb points towards the direction d

The direction of the force is,

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

Thus, the direction of the magnetic force is $\left(\frac{1}{\sqrt{2}}(\widehat{i}+\widehat{k})\right)$.

Conclusion:

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

To determine

How to find the force exerted on the forth segment from the forces on the other three.

Answer to Problem 28P

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

Explanation of Solution

Given info: The length of each edge of the cube is $40.0\text{cm}$, the electric current is $5.00\text{A}$ and the magnitude of magnetic field is $0.020\text{T}$

By the parallelogram law of forces, when the forces on three of the arms of a parallelogram are provided then the magnitude of force on the forth arm is equal to the resultant of the other three forces.

According to the parallelogram law of vectors,

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

Here,

${\overrightarrow{F}}_4$ is the force of the resultant.

${\overrightarrow{F}}_3$ is the force on the third arm of the parallelogram.

${\overrightarrow{F}}_1$ is the force on the first arm of the parallelogram.

${\overrightarrow{F}}_2$ is the force on the second arm of the parallelogram.

Conclusion:

Therefore, the force exerted on the forth segment from the forces on the other three can be calculated by the parallelogram law of vectors.