Chapter 29, Problem 52P

(a)

To determine

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

Answer to Problem 52P

The direction of magnetic force exerted on wire segment $ab$ is along the positive $x$ axis.

Explanation of Solution

Write the expression to calculate the magnetic force exerted on $ab$.

    $\overrightarrow{F_{\text{ab}}}=I\overrightarrow{L}\times \overrightarrow{B}$                                                                                                               (I)

Here, $\overrightarrow{F_{\text{ab}}}$ is the magnetic force exerted on $ab$, $I$ is the current, $\overrightarrow{B}$ is the magnetic field and $\overrightarrow{L}$ is the length of the segment $ab$.

Write the expression to calculate the resultant magnetic field.

    $\overrightarrow{B}=\left(B\cos \theta \right)\widehat{\text{j}}+\left(B\sin \theta \right)\widehat{\text{k}}$                                                                                        (II)

Here, $\overrightarrow{B}$ is the resultant magnetic field and $\theta$ is the angle made by the resultant magnetic field with the $y$ axis.

Conclusion:

Substitute $1.50\text{T}$ for $B$ and $40°$ for $\theta$ in equation (II) to solve for $\overrightarrow{B}$.

    $\begin{array}{c}\overrightarrow{B}=\left(1.50\text{T}\cos 40°\right)\widehat{\text{j}}+\left(1.50\text{T}\sin 40°\right)\widehat{\text{k}}\\ =\left(1.150\widehat{\text{j}}\right)\text{T}+\left(0.96\widehat{\text{k}}\right)\text{T}\end{array}$

Substitute $\left(1.150\widehat{\text{j}}\right)\text{T}+\left(0.96\widehat{\text{k}}\right)\text{T}$ for $\overrightarrow{B}$, $\left(0.500\text{m}\right)\widehat{\text{j}}$ for $\overrightarrow{L}$ and $0.900\text{A}$ for $I$ in equation (I) to solve for $\overrightarrow{F_{\text{ab}}}$.

    $\begin{array}{c}\overrightarrow{F_{\text{cd}}}=\left(0.900\text{A}\right)\left(\left(0.500\widehat{\text{j}}\right)\text{m}\right)\times \left(\left(1.150\widehat{\text{j}}\right)\text{T}+\left(0.96\widehat{\text{k}}\right)\text{T}\right)\\ =\left(0.900\text{A}\right)\left[\left(\left(0.500\widehat{\text{j}}\right)\text{m}\times \left(1.150\widehat{\text{j}}\right)\text{T}\right)+\left(\left(0.500\widehat{\text{j}}\right)\text{m}\times \left(0.96\widehat{\text{k}}\right)\text{T}\right)\right]\\ =\left(0.900\text{A}\right)\left[0+\left(0.48\widehat{\text{i}}\right)\text{m}\right]\\ =\left(0.432\widehat{\text{i}}\right)\text{N}\end{array}$

Therefore, the direction of magnetic force exerted on wire segment $ab$ is along the positive $x$ axis.

(b)

To determine

The direction of torque associated with this force.

Answer to Problem 52P

The direction of torque associated with this force is along the negative $z$ axis

Explanation of Solution

Write the expression to calculate the torque.

    $\overrightarrow{\tau }=\overrightarrow{r}+\overrightarrow{F_{\text{ab}}}$                                                                                                                (III)

Here, $\overrightarrow{\tau }$ is the torque associated with the magnetic force and $\overrightarrow{r}$ is the distance from the axis.

Conclusion:

Substitute $\left(0.500\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{r}$ and $\left(0.432\widehat{\text{i}}\right)\text{N}$ for $\overrightarrow{F_{\text{ab}}}$ in equation (III) to solve for $\overrightarrow{\tau }$.

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

Therefore, the direction of torque associated with this force is along the negative $z$ axis.

(c)

To determine

Write the expression to calculate the magnetic force exerted on $cd$.

Answer to Problem 52P

The direction of magnetic force exerted on wire segment $cd$ is along the negative $x$ axis.

Explanation of Solution

Write the expression to calculate the magnetic force exerted on $cd$.

    $\overrightarrow{F_{\text{cd}}}=I\overrightarrow{L}\times \overrightarrow{B}$                                                                                                            (IV)

Here, $\overrightarrow{F_{\text{cd}}}$ is the magnetic force exerted on $cd$, $I$ is the current, $\overrightarrow{B}$ is the magnetic field and $\overrightarrow{L}$ is the length of the segment $ab$.

Conclusion:

Substitute $\left(1.150\text{T}\right)\widehat{\text{j}}+\left(0.96\text{T}\right)\widehat{\text{k}}$ for $\overrightarrow{B}$, $\left(-0.500\text{m}\right)\widehat{\text{j}}$ for $\overrightarrow{L}$ and $0.900\text{A}$ for $I$ in equation (IV) to solve for $\overrightarrow{F_{\text{ab}}}$.

    $\begin{array}{c}\overrightarrow{F_{\text{cd}}}=\left(0.900\text{A}\right)\left(\left(-0.500\widehat{\text{j}}\right)\text{m}\right)\times \left(\left(1.150\widehat{\text{j}}\right)\text{T}+\left(0.96\widehat{\text{k}}\right)\text{T}\right)\\ =\left(0.900\text{A}\right)\left[\left(\left(-0.500\widehat{\text{j}}\right)\text{m}\times \left(1.150\widehat{\text{j}}\right)\text{T}\right)+\left(\left(-0.500\widehat{\text{j}}\right)\text{m}\times \left(0.96\widehat{\text{k}}\right)\text{T}\right)\right]\\ =\left(0.900\text{A}\right)\left[0+\left(-0.48\widehat{\text{i}}\right)\text{m}\right]\\ =\left(-0.432\widehat{\text{i}}\right)\text{N}\end{array}$

Therefore, the direction of magnetic force exerted on wire segment $cd$ is along the negative $x$ axis.

(d)

To determine

The direction of torque associated with this force.

Answer to Problem 52P

The direction of torque associated with this force is along the negative $z$ axis

Explanation of Solution

Write the expression to calculate the torque.

    $\overrightarrow{\tau }=\overrightarrow{r}+\overrightarrow{F_{\text{ab}}}$                                                                                                                (V)

Here, $\overrightarrow{\tau }$ is the torque associated with the magnetic force and $\overrightarrow{r}$ is the distance from the axis.

Conclusion:

Substitute $\left(-0.500\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{r}$ and $\left(-0.432\widehat{\text{i}}\right)\text{N}$ for $\overrightarrow{F_{\text{ab}}}$ in equation (V) to solve for $\overrightarrow{\tau }$.

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

Therefore, the direction of torque associated with this force is along the negative $z$ axis.

(e)

To determine

Whether the obtained forces combine to rotate the loop along the $x$ axis.

Answer to Problem 52P

The obtained forces cannot combine to rotate the loop along the $x$ axis because the resultant force is zero when both forces are combined.

Explanation of Solution

Since, the magnitude of magnetic forces on the segments $ab$ and $cd$ are equal but they  act in opposite directions.

Hence, the net force on combining both the forces, the resultant force becomes zero.

Thus, the magnetic forces cannot rotate the loop along the $x$ axis.

Therefore, the obtained forces cannot combine to rotate the loop along the $x$ axis.

(f)

To determine

Whether the obtained forces can affect the motion of loop in anyway.

Answer to Problem 52P

The obtained forces cannot affect the motion of loop in anyway

Explanation of Solution

Since, the magnetic field, the current and the length is constant, Hence, the magnetic force obtained will be constant.

So, the magnetic forces will be able to rotate the loop only and shall not affect the motion of the loop in anyway.

Therefore, the obtained forces cannot affect the motion of loop in anyway

(g)

To determine

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

Answer to Problem 52P

The direction of magnetic force exerted on wire segment $bc$ is along the $y\text{-}z$ plane.

Explanation of Solution

Write the expression to calculate the magnetic force exerted on $bc$.

    $\overrightarrow{F_{\text{bc}}}=I\overrightarrow{L}\times \overrightarrow{B}$                                                                                                            (VI)

Here, $\overrightarrow{F_{\text{bc}}}$ is the magnetic force exerted on $bc$, $I$ is the current, $\overrightarrow{B}$ is the magnetic field and $\overrightarrow{L}$ is the length of the segment $ab$.

Conclusion:

Substitute $\left(1.150\widehat{\text{j}}\right)\text{T}+\left(0.96\widehat{\text{k}}\right)\text{T}$ for $\overrightarrow{B}$, $\left(0.300\text{m}\right)\widehat{\text{i}}$ for $\overrightarrow{L}$ and $0.900\text{A}$ for $I$ in equation (VI) to solve for $\overrightarrow{F_{\text{bc}}}$.

    $\begin{array}{c}\overrightarrow{F_{\text{bc}}}=\left(0.900\text{A}\right)\left(\left(0.300\widehat{\text{i}}\right)\text{m}\right)\times \left(\left(1.150\widehat{\text{j}}\right)\text{T}+\left(0.96\widehat{\text{k}}\right)\text{T}\right)\\ =\left(0.900\text{A}\right)\left[\left(\left(0.300\widehat{\text{i}}\right)\text{m}\times \left(1.150\widehat{\text{j}}\right)\text{T}\right)+\left(\left(0.300\widehat{\text{i}}\right)\text{m}\times \left(0.96\widehat{\text{k}}\right)\text{T}\right)\right]\\ =\left(0.900\text{A}\right)\left[\left(0.345\left(\widehat{\text{i}}\text{×}\widehat{\text{j}}\right)\right)\text{T-m}+\left(0.288\widehat{\text{j}}\right)\text{T-m}\right]\\ =\left(0.31\widehat{\text{k}}\right)\text{N}-\left(0.26\widehat{\text{j}}\right)\text{N}\end{array}$

Therefore, the direction of magnetic force exerted on wire segment $bc$ is along the $y\text{-}z$ plane.

(h)

To determine

The direction of torque associated with this force.

Answer to Problem 52P

The direction of torque associated with this force is along the negative $y$ axis

Explanation of Solution

Write the expression to calculate the torque.

    $\overrightarrow{\tau }=\overrightarrow{r}+\overrightarrow{F_{\text{bc}}}$                                                                                                              (VII)

Here, $\overrightarrow{\tau }$ is the torque associated with the magnetic force and $\overrightarrow{r}$ is the distance from the axis.

Conclusion:

Substitute $\left(0.300\widehat{\text{i}}\right)\text{m}$ for $\overrightarrow{r}$ and $\left(0.31\widehat{\text{k}}\right)\text{N}-\left(0.26\widehat{\text{j}}\right)\text{N}$ for $\overrightarrow{F_{\text{bc}}}$ in equation (VII) to solve for $\overrightarrow{\tau }$.

    $\begin{array}{c}\overrightarrow{\tau }=\left(0.300\widehat{\text{i}}\right)\text{m}\left(\left(0.31\widehat{\text{k}}\right)\text{N}-\left(0.26\widehat{\text{j}}\right)\text{N}\right)\\ =\left(-0.093\widehat{\text{j}}-0.078\widehat{\text{j}}\right)\text{N-m}\\ \text{=}\left(-0.171\widehat{\text{j}}\right)\text{N-m}\end{array}$

Therefore, the direction of torque associated with this force is along the negative $y$ axis.

(i)

To determine

The direction of torque on segment $ad$.

Answer to Problem 52P

The direction of the torque associated with segment cannot be defined because the torque on segment $ad$ is zero

Explanation of Solution

Since, the lever arm on the segment $ad$ is zero. Hence, the torque on this segment will be zero.

Therefore, the direction of the torque associated with segment cannot be defined because the torque on segment $ad$ is zero.

(j)

To determine

The direction of rotation of the loop when it is released from rest.

Answer to Problem 52P

The rotation of loop will be in counter clockwise direction about the $x$ axis.

Explanation of Solution

The direction of magnetic force exerted on wire segment $ab$ is along the positive $x$ axis

The direction of torque on wire segment $ab$ is along the negative $z$ axis

The direction of magnetic force exerted on wire segment $cd$ is along the negative $x$ axis

The direction of torque on wire segment $cd$ is along the negative $z$ axis

This signifies that the torque on segment $ab$ and $cd$ is in negative $z$ direction about $x$ axis.

Therefore, the rotation of loop will be in counter clockwise direction about the $x$ axis.

(k)

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 52P

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

Explanation of Solution

Write the expression to calculate the area of the loop.

    $A=l\times b$                                                                                                                (VIII)

Here, $A$ is the area of the loop, $l$ is the length and $b$ is the breadth.

Write the expression to calculate the magnetic moment of the loop.

    $\mu =NIA$                                                                                                                  (IX)

Here, $\mu$ is the magnetic moment of the loop, $N$ is the number of turns and $I$ is the current through the loop.

Substitute $\left(l\times b\right)$ for $A$ in equation (II).

    $\mu =NI\left(l\times b\right)$                                                                                                            (X)

Conclusion:

Substitute $0.500\text{m}$ for $l$, $0.300\text{m}$ for $b$, $1$ for $N$ and $0.900\text{A}$ for $I$ in equation (X) to solve for $\mu$.

    $\begin{array}{c}\mu =\left(1\right)\left(0.900\text{A}\right)\left(0.500\text{m}\times 0.300\text{m}\right)\\ =0.135{\text{A-m}}^{\text{2}}\end{array}$

Therefore, the magnitude of the magnetic moment of the loop is $0.135{\text{A-m}}^{\text{2}}$.

(l)

To determine

The angle between the magnetic moment vector and the magnetic field.

Answer to Problem 52P

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

Explanation of Solution

The current flowing through the loop is in clockwise direction. So by the right hand  thumb rule, the direction of magnetic field will be in the downward direction along the negative $z$ direction.

So the angle between the magnetic moment vector and the magnetic field will be,

    $\varphi =90°+\theta$                                                                                                              (XI)

Here, $\varphi$ is the angle between the magnetic moment vector and the magnetic field

Conclusion:

Substitute $40°$ for $\theta$ in equation (XI) to solve for $\varphi$.

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

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

(l)

To determine

The torque in the loop.

Answer to Problem 52P

The torque in the loop is $0.155\text{N-m}$.

Explanation of Solution

Write the expression to calculate the torque on the loop.

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

Here, $\tau$ is the torque through the loop.

Conclusion:

Substitute $0.135{\text{A-m}}^2$ for $\mu$, $1.5\text{T}$ for $B$ and $130°$ for $\varphi$ in equation (XII) to solve for $\tau$.

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

Therefore, the torque in the loop is $0.155\text{N-m}$.