Chapter 22, Problem 26P

(a)

To determine

The direction of the magnetic force exerted on wire segment $ab$ .

Answer to Problem 26P

The direction of the force is in the positive $x$ axis.

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment $ab$, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive $x$ axis.

Conclusion:

Therefore, the direction of the force is in the positive $x$ axis.

(b)

To determine

The direction of the torque associated with the above force about the axis through origin.

Answer to Problem 26P

The direction of the torque associated with the force on the segment $ab$ about an axis through the origin is in the direction of negative $z$ axis.

Explanation of Solution

Write the expression for torque associated with the force,

  $\overrightarrow{\tau }=\overrightarrow{r}\times {\overrightarrow{F}}_{ab}$        (I)

Here, $\overrightarrow{\tau }$ is the torque, ${\overrightarrow{F}}_{ab}$  is the force on the segment $ab$, and $\overrightarrow{r}$  is the distance from the axis.

Conclusion:

Substitute $\left(0.500\text{m}\right)\widehat{\text{j}}$ for $\overrightarrow{r}$ , $0.432\text{N}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{ab}$  equation (I).

    $\begin{array}{c}\overrightarrow{\tau }=\left(0.500\text{m}\right)\widehat{\text{j}}\times \left(0.432\text{N}\right)\widehat{\text{i}}\\ \text{=}\left(0.500\text{m}\right)\left(0.432\text{N}\right)\left(\widehat{\text{j}}\times \widehat{\text{i}}\right)\\ =-0.216\text{Nm}\widehat{\text{k}}\end{array}$

The torque related to this expression is negative of unit vector $k$. Thus the toque associated with the force on a segment  $ab$ is in the direction of the negative $z$ axis.

(c)

To determine

The direction of the magnetic force exerted on wire segment $cd$ .

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative $x$ axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

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

Here, $F$ is the force, $L$ is the distance from the axis, $I$ is the current of the segment $cd$,  and  $B$ is the magnetic field.

Conclusion:

Substitute $\left(1.15\text{T}\right)\widehat{\text{j}}\text{ +}\left(0.96\text{T}\right)\widehat{\text{k}}$ for $B$ , $0.009\text{A}$ for $I$ , and $\left(-0.500\text{m}\right)\text{j}$ for $\overrightarrow{L}$ in the above equation (I) to find the value of $F$ .

    $\begin{array}{c}\overrightarrow{F}=0.009\text{A}\times \left(\left(\left(-0.500\text{m}\right)\widehat{\text{j}}\right)\times \left(1.15\text{T}\right)\widehat{\text{j}}\text{ +}\left(0.96\text{T}\right)\widehat{\text{k}}\right)\\ =\left(0.009\text{A}\right)\left(0.500\text{m}\right)\left(1.15\text{T}\right)\left(\widehat{\text{j}}\text{×}\widehat{\text{j}}\right)-\left(0.009\text{A}\right)\left(0.500\text{m}\right)\left(0.96\text{T}\right)\left(\widehat{\text{j}}\text{×}\widehat{\text{k}}\right)\\ =-0.432\text{N}\widehat{\text{i}}\end{array}$

From the above expression it can understood that the magnetic force in the negative of the unit vector $i$ . Thus, the direction of the magnetic force excreted on the segment is in the negative $x$ axis.

(d)

To determine

The direction of the torque associated with the force on the segment $cd$ about an axis through the origin.

Answer to Problem 26P

The direction of the torque associated with the force on the segment $cd$ about an axis through the origin is in the direction along the negative $z$ axis.

Explanation of Solution

Write the expression for torque associated with the force,

  $\overrightarrow{\tau }=\overrightarrow{r}\times {\overrightarrow{F}}_{cd}$        (I)

Here, $\overrightarrow{\tau }$ is the torque, ${\overrightarrow{F}}_{cd}$  is the force on the segment $cd$, and $\overrightarrow{r}$  is the distance from the axis.

Conclusion:

Substitute $\left(0.500\text{m}\right)\widehat{\text{j}}$ for $\overrightarrow{r}$ , $0.432\text{N}\widehat{\text{i}}$ for ${\overrightarrow{F}}_{cd}$  equation (I).

    $\begin{array}{c}\overrightarrow{\tau }=\left(0.500\text{m}\right)\widehat{\text{j}}\times \left(0.432\text{N}\right)\widehat{\text{i}}\\ \text{=}\left(0.500\text{m}\right)\left(0.432\text{N}\right)\left(\widehat{\text{j}}\times \widehat{\text{i}}\right)\\ =-0.216\text{Nm}\widehat{\text{k}}\end{array}$

The torque related to this expression is negative of unit vector $k$. Thus the toque associated with the force on a segment  $cd$ is in the direction of the negative $z$ axis.

(e)

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the $x$ axis.

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment $ab$  is along the positive $x$ axis and the force examined is on the segment $cd$ which is along negative $x$ axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment $ab$ cancel the effect on the force on the segment $cd$.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

(f)

To determine

Whether the force affect the motion of the loop.

Answer to Problem 26P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Write the expression for the magnetic field in a wire,

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

Here, $F$ is the force, $L$ is the distance from the axis, $I$ is the current of the segment $cd$,  and  $B$ is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, the magnetic field, the current, and the length of the loop is constant, therefore the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

(g)

To determine

The direction of the magnetic force exerted on the segment $bc$ .

Answer to Problem 26P

The magnetic force will on segment $bc$ is in the direction along $yz$ plane.

Explanation of Solution

Write the expression for the magnetic field in a wire,

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

Here, $F$ is the force, $L$ is the distance from the axis, $I$ is the current of the segment $cd$,  and  $B$ is the magnetic field.

Conclusion:

Substitute $\left(1.15\text{T}\right)\widehat{\text{j}}\text{ +}\left(0.96\text{T}\right)\widehat{\text{k}}$ for $B$ , $0.009\text{A}$ for $I$ , and $\left(0.300\text{m}\right)\text{j}$ for $\overrightarrow{L}$ in the above equation (I) to find the value of $F$ .

    $\begin{array}{c}\overrightarrow{F}=0.009\text{A}\times \left(\left(\left(\text{0}\text{.300m}\right)\widehat{\text{i}}\right)\times \left(1.15\text{T}\right)\widehat{\text{j}}\text{ +}\left(0.96\text{T}\right)\widehat{\text{k}}\right)\\ =\left(0.009\text{A}\right)\left(0.300\text{m}\right)\left(1.15\text{T}\right)\left(\widehat{\text{i}}\text{×}\widehat{\text{j}}\right)-\left(0.009\text{A}\right)\left(0.500\text{m}\right)\left(0.96\text{T}\right)\left(\widehat{\text{i}}\text{×}\widehat{\text{k}}\right)\\ =-\left(0.26\text{N}\right)\widehat{\text{j}}\text{+}\left(0.16\text{N}\right)\widehat{\text{k}}\end{array}$

From the above expression it can understood that the magnetic force in the negative of the unit vector $j$ . Thus, the direction of the magnetic force excreted on the segment $bc$ is in the direction along $yz$ plane.

Conclusion:

The magnetic force will on segment $bc$ is in the direction along $yz$ plane.

(h)

To determine

The direction of the torque associated with the force on the segment $bc$ about the axis through the origin.

Answer to Problem 26P

The direction of the torque associated with the force on the segment $bc$ about an axis through the origin is in the direction along the positive $x$ axis.

Explanation of Solution

Write the expression for torque associated with the force,

  $\overrightarrow{\tau }=\overrightarrow{r}\times {\overrightarrow{F}}_{bc}$        (I)

Here, $\overrightarrow{\tau }$ is the torque, ${\overrightarrow{F}}_{bc}$  is the force on the segment $bc$, and $\overrightarrow{r}$  is the distance from the axis.

Conclusion:

Substitute $\left(0.300\text{m}\right)\widehat{i}$ for $\overrightarrow{r}$ , $-\left(0.26\text{N}\right)\widehat{\text{j}}\text{+}\left(0.16\text{N}\right)\widehat{\text{k}}$ for ${\overrightarrow{F}}_{bc}$  equation (I).

    $\begin{array}{c}\overrightarrow{\tau }=\left(0.300\text{m}\right)\widehat{\text{j}}\times \left(-\left(0.26\text{N}\right)\widehat{\text{j}}\text{+}\left(0.16\text{N}\right)\widehat{\text{k}}\right)\\ \text{=}\left(0.300\text{m}\right)\left(0.26\text{N}\right)\left(\widehat{\text{j}}\text{×}\widehat{\text{j}}\right)+\left(0.300\text{m}\right)\left(0.16\text{N}\right)\left(\widehat{\text{j}}\times \widehat{\text{k}}\right)\\ =0.048\text{N}\times \text{m}\widehat{\text{i}}\end{array}$

The torque related to this expression is positive of unit vector $\widehat{\text{i}}$. Thus the toque associated with the force on a segment  $bc$ is in the direction of the positive $x$ axis.

(i)

To determine

The direction of the torque on the segment $ad$ about the axis through the origin.

Answer to Problem 26P

The direction of the torque associated segment $ad$ is zero

Explanation of Solution

Write the expression for torque associated with the force,

  $\overrightarrow{\tau }=\overrightarrow{r}\times {\overrightarrow{F}}_{ad}$        (I)

Here, $\overrightarrow{\tau }$ is the torque, ${\overrightarrow{F}}_{ad}$  is the force on the segment $ad$, and $\overrightarrow{r}$  is the distance from the axis.

Conclusion:

The segment $ad$ is turns along the $x$ axis. This makes the distance of the segment $ad$ from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment $ad$ is zero

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment $ab$ and $cd$ is along the negative $z$ direction and the torque on the segment $ad$ is zero.

Thus, the net torque is in the direction along the positive $z$ axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

(k)

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 26P

The magnitude of the magnetic moment of the loop is $0.135\text{A}\cdot {\text{m}}^{\text{2}}$ .

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

  $\mu =NIlb$        (I)

Here, $N$ is the number of turns, $I$ is the current , $l$ is the length of the loop, $b$ is the breadth of the loop.

Conclusion:

Substitute $1$  for $N$ , $0.900\text{A}$ for $I$ , $\text{0}\text{.500m}$ for  $l$, and $0.300\text{m}$ for $b$ in the expression (I).

    $\begin{array}{c}\mu =1\left(0.900\text{A}\right)\left(\text{0}\text{.500m}\right)\left(0.300\text{m}\right)\\ =0.135\text{A}\cdot {\text{m}}^{\text{2}}\end{array}$

The magnitude of the magnetic moment of the loop is $0.135\text{A}\cdot {\text{m}}^{\text{2}}$ .

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 26P

The angle between the magnetic moment vector and magnetic field is $130°$ .

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative $z$ direction and the angle between the magnetic moment and the magnetic field is,

  $\begin{array}{c}\varphi =90°+40°\\ =130°\end{array}$

Conclusion:

Therefore, the angle between the magnetic moment vector and magnetic field is $130°$ .

(m)

To determine

The torque on the loop using the results of part (k) and (l)

Answer to Problem 26P

The torque on the loop is $0.155\text{N}\cdot \text{m}$ .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

  $\tau =\mu B\sin \varphi$        (I)

Here, $\tau$ is the torque, $\mu$ is the permeability, and $B$ is the magnetic field

Conclusion:

Substitute $0.135\text{A}\cdot {\text{m}}^2$ for $\mu$ , $1.5\text{T}$ for $B$ and $130°$ for $\varphi$ in the above expression.

    $\begin{array}{c}\tau =\left(0.135\text{A}\cdot {\text{m}}^2\right)\left(1.5\text{T}\right)\sin \left(130°\right)\\ =0.155\text{N}\cdot \text{m}\end{array}$

The torque on the loop is $0.155\text{N}\cdot \text{m}$ .