Chapter 28, Problem 34P

A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l).

Figure P28.34

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment $ab$, an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

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

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The Formula to calculate the torque associated with the force is,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F_{\text{ab}}}$ (1)

Here,

$\overrightarrow{\tau }$ is the associated torque.

$\overrightarrow{F_{\text{ab}}}$ is the force on segment $ab$.

$\overrightarrow{r}$ is the distance from the axis.

Substitute $\left(0.500\text{m}\right)\hat{\text{j}}$ for $\overrightarrow{r}$ and $\left(0.432\text{N}\right)\hat{\text{i}}$ for $\overrightarrow{F_{\text{ab}}}$ in the equation (1).

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

The torque associated in the above expression in the negative of the unit vector $\hat{\text{k}}$. Therefore, the torque associated with the force on segment $ab$ is in the direction of negative z-axis.

Conclusion:

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

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The Formula to calculate the magnetic force on the wire segment $cd$ is,

$\overrightarrow{F_{\text{cd}}}=I\overrightarrow{L}\times \overrightarrow{B}$ (2)

Here,

$I$ is the current on segment $cd$.

$\overrightarrow{F_{\text{cd}}}$ is the force on segment $cd$.

$\overrightarrow{L}$ is the distance from the axis.

$\overrightarrow{B}$ is the magnitude of magnetic field.

Substitute $0.900\text{A}$ for $I$, $\left(-0.500\text{m}\right)\hat{\text{j}}$ for $\overrightarrow{L}$ and $\left(1.15\text{T}\right)\hat{\text{j}}+\left(0.96\text{T}\right)\hat{\text{k}}$ for $\overrightarrow{B}$ in the equation (2).

$\begin{array}{c}\overrightarrow{F_{\text{cd}}}=\left(0.900\text{A}\right)\times \left\{\left[-\left(0.500\text{m}\right)\hat{\text{j}}\right]\times \left[\left(1.15\text{T}\right)\hat{\text{j}}+\left(0.96\text{T}\right)\hat{\text{k}}\right]\right\}\\ =\left(0.900\text{A}\right)\left(0.500\text{m}\right)\left(1.15\text{T}\right)\left(\hat{\text{j}}\times \hat{\text{j}}\right)-\left(0.900\text{A}\right)\left(0.500\text{m}\right)\left(0.96\text{T}\right)\left(\hat{\text{j}}\times \hat{\text{k}}\right)\\ =0-\left(0.900\text{A}\right)\left(0.500\text{m}\right)\left(0.96\text{T}\right)\left(\hat{\text{j}}\times \hat{\text{k}}\right)\\ =-\left(0.432\text{N}\right)\hat{\text{i}}\end{array}$

In the above expression it is shown that the magnetic force is in the negative of the unit vector $\hat{\text{i}}$. Therefore, the magnetic force on segment $cd$ is in the direction of negative x-axis.

Conclusion:

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

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The Formula to calculate the torque associated with the force is,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F_{\text{cd}}}$ (3)

Here,

$\overrightarrow{\tau }$ is the associated torque.

$\overrightarrow{F_{\text{cd}}}$ is the force on segment $cd$.

$\overrightarrow{r}$ is the distance from the axis.

Substitute $\left(-0.500\text{m}\right)\hat{\text{j}}$ for $\overrightarrow{r}$ and $\left(-0.432\text{N}\right)\hat{\text{i}}$ for $\overrightarrow{F_{\text{cd}}}$ in the equation (3).

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

The torque associated in the above expression is in the negative of the unit vector $\hat{\text{k}}$. Therefore, the torque associated with the force on segment $cd$ is along the direction of negative z-axis.

Conclusion:

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

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 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

As the force examined on the segment $ab$ is along the positive x-axis and the force examined on the segment $cd$ is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment $ab$ cancel the effect of the force examined on the segment $cd$.

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem 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.

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The Formula to calculate the magnetic force on the wire segment $bc$ is,

$\overrightarrow{F_{\text{bc}}}=I\overrightarrow{L}\times \overrightarrow{B}$ (4)

Here,

$\overrightarrow{F_{\text{bc}}}$ is the force on segment $cd$.

Substitute $0.900\text{A}$ for $I$, $\left(0.300\text{m}\right)\hat{\text{j}}$ for $\overrightarrow{L}$ and $\left(1.15\text{T}\right)\hat{\text{j}}+\left(0.96\text{T}\right)\hat{\text{k}}$ for $\overrightarrow{B}$ in the equation (2).

$\begin{array}{c}\overrightarrow{F_{\text{bc}}}=\left(0.900\text{A}\right)\times \left\{\left[\left(0.300\text{m}\right)\hat{\text{i}}\right]\times \left[\left(1.15\text{T}\right)\hat{\text{j}}+\left(0.96\text{T}\right)\hat{\text{k}}\right]\right\}\\ =\left(0.900\text{A}\right)\left(0.300\text{m}\right)\left(1.15\text{T}\right)\left(\hat{\text{i}}\times \hat{\text{j}}\right)-\left(0.900\text{A}\right)\left(0.500\text{m}\right)\left(0.96\text{T}\right)\left(\hat{\text{i}}\times \hat{\text{k}}\right)\\ =\left(0.16\text{N}\right)\hat{\text{k}}-\left(0.26\text{N}\right)\left(\hat{\text{j}}\right)\\ =-\left(0.26\text{N}\right)\hat{\text{j}}+\left(0.16\text{N}\right)\hat{\text{k}}\end{array}$

In the above expression it is shown that the magnetic force is in the negative of the unit vector $\hat{\text{j}}$ and positive of the unit vector $\hat{\text{j}}$. Therefore, the magnetic force on segment $bc$ is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment $bc$ is in the direction along yz-plane.

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The Formula to calculate the torque associated with the force is,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F_{bc}}$

Here,

$\overrightarrow{F_{bc}}$ is the force on segment $bc$.

Substitute $\left(0.300\text{m}\right)\hat{\text{i}}$ for $\overrightarrow{r}$ and $-\left(0.26\text{N}\right)\hat{\text{j}}+\left(0.16\text{N}\right)\hat{\text{k}}$ for $\overrightarrow{F_{\text{bc}}}$ in the above expression.

$\begin{array}{c}\overrightarrow{\tau }=\left(0.300\text{m}\right)\hat{\text{j}}\times \left[-\left(0.26\text{N}\right)\hat{\text{j}}+\left(0.16\text{N}\right)\hat{\text{k}}\right]\\ =-\left(0.300\text{m}\right)\left(0.26\text{N}\right)\left(\hat{\text{j}}\times \hat{\text{j}}\right)+\left(0.300\text{m}\right)\left(0.16\text{N}\right)\left(\hat{\text{j}}\times \hat{\text{k}}\right)\\ =0+\left(0.048\text{N}\cdot \text{m}\right)\hat{\text{i}}\\ =0.048\text{N×m}\hat{\text{i}}\end{array}$

The torque associated in the above expression is in the positive of the unit vector $\hat{\text{i}}$. Therefore, the torque associated with the force on segment $bc$ is along the direction of positive x-axis.

Conclusion:

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

To determine

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

Answer to Problem 34P

The torque on the segment $ad$ is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The Formula to calculate the torque associated with the force is,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F_{ad}}$

The segment $ad$ is pivoted along the x-axis. This makes the distance of the segment $ad$ from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment $ad$ is zero.

Conclusion:

Therefore, the torque on the segment $ad$ is zero.

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

As the torque on the segment $ab$ and $cd$ is along the negative z-direction and the torque on segment $ad$ is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is $\text{0}\text{.135}\text{A}\cdot {\text{m}}^2$.

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

The formula to calculate the magnetic moment of the loop is,

$\mu =NIlb$

Here,

$N$ is the number of turns.

Substitute $1$ for $N$, $0.900\text{A}$ for $I$, $0.500\text{m}$ for $l$ and $0.300\text{m}$ for $b$ in above equation.

$\begin{array}{c}\mu =1\times 0.900\text{A}\times 0.500\text{m}\times 0.300\text{m}\\ =\text{0}\text{.135}\text{A}\cdot {\text{m}}^2\end{array}$

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is $\text{0}\text{.135}\text{A}\cdot {\text{m}}^2$.

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

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

Conclusion:

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

To determine

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

Answer to Problem 34P

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

Explanation of Solution

Given Info: The length of the rectangular loop is $0.500\text{m}$, the breadth of the rectangular loop is $0.300\text{m}$, the magnitude of the magnetic field is $1.5\text{T}$, the current in the rectangular loop is $0.900\text{A}$, the directed angle of the magnetic field with respect to the y-axis is $40.0°$.

Formula to calculate the torque in current carrying wire is,

$\tau =\mu B\sin \varphi$

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

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

Conclusion:

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