Chapter 19, Problem 42P

A rectangular loop has dimensions 0.500 m by 0.300 m. The loop is hinged along the x-axis and lies in the xy-plane (Fig. P19.42). A uniform magnetic field of 1.50 T is directed at an angle of 40.0° with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) (a) In what direction is magnetic force exerted on wire segment ab? What is the direction of the magnetic torque associated with this force, as computed with respect to the x-axis? (b) What is the direction of the magnetic force exerted on segment cd? What is the direction of the magnetic torque associated with this force, again computed with respect to the x-axis? (c) Can the forces examined in parts (a) and (b) combine to cause the loop to rotate around the x-axis? Can they affect the motion of the loop in any way? Explain. (d) What is the direction (in the yz-plane) of the magnetic force exerted on segment bc? Measuring torques with respect to the x-axis, what is the direction of the torque exerted by the force on segment bc? (e) Looking toward the origin along the positive x-axis. Will the loop rotate clockwise or counterclockwise? (f) Compute the magnitude of the magnetic moment of the loop. (g) What is the angle between the magnetic moment vector and the magnetic field? (h) Compute the torque on the loop using the values found for the magnetic moment and magnetic field.

Figure P19.42

To determine

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

Answer to Problem 42P

The direction of the force on the segment ab will direct in the positive x-direction.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

The direction can be determined by right hand rule 1.

Right hand rule 1: When the fingers of the right hand point in the direction of current and curl towards the magnetic field, the thumb points in the direction of the force.

The current in the segment ab is flowing in the positive y-direction. The magnetic field lies parallel to the yz-plane. By the application of the right hand rule 1 the magnetic force on the segment ab will direct in the positive x-direction.

Conclusion: The direction of the force on the segment ab will direct in the positive x-direction.

To determine

The direction of magnetic torque exerted on the wire segment $ab$ computed with respect to the x-axis.

Answer to Problem 42P

The torque about x-axis due to the magnetic force will be zero.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

Since the force is parallel to the x-axis, the torque with respect to x-axis will be zero.

Conclusion: The torque about x-axis due to the magnetic force will be zero.

To determine

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

Answer to Problem 42P

The direction of the force on the segment $cd$ will direct in the negative x-direction.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

The direction of the magnetic force can be determined by right hand rule 1.

Right hand rule 1: When the fingers of the right hand point in the direction of current and curl towards the magnetic field, the thumb points in the direction of the force.

The current in the segment $cd$ is flowing in the negative y-direction. The magnetic field lies parallel to the yz-plane. By the application of the right hand rule 1 the magnetic force on the segment $cd$ will direct in the negative x-direction.

Conclusion: The direction of the force on the segment $cd$ will direct in the negative x-direction.

To determine

The direction of magnetic torque exerted on the wire segment $cd$ computed with respect to the x-axis.

Answer to Problem 42P

The torque about x-axis due to the magnetic force will be zero.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

Since the force is along negative x-axis hence parallel to the -axis, the torque with respect to x-axis will be zero. x

Conclusion: The torque about x-axis due to the magnetic force will be zero.

To determine

Whether the magnetic forces on segment ab and cd can combine to rotate the loop.

Answer to Problem 42P

The magnetic forces on segment ab and cd cannot combine to rotate the loop.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$  and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$ .

Since the loop is hinged on the x-axis, if the loop has to rotate, it has to rotate about x-axis. From section (a) and (b), the forces on segment ab and cd are parallel to the rotation axis that is x-axis. Hence they don’t create any torque about the x-axis. Hence they cannot combine to rotate the loop.

Conclusion: The magnetic forces on segment ab and cd cannot combine to rotate the loop

To determine

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

Answer to Problem 42P

The magnetic force on the segment $bc$ will be at $130°$ counterclockwise from the positive y-axis in the yz-plane.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field i $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

The direction of the magnetic force can be determined by right hand rule 1.

Right hand rule 1: When the fingers of the right hand point in the direction of current and curl towards the magnetic field, the thumb points in the direction of the force.

The current in the segment $bc$ is flowing in the positive x-direction. The magnetic field lies parallel to the yz-plane.

Since the magnetic force involves the cross product of the direction of current and the magnetic field, the magnetic force will be perpendicular to x-axis as well as it will be perpendicular to the magnetic field direction.

Thus, the magnetic force will be in the yz-plane. By the application of the right hand rule 1 the magnetic force on the segment $bc$ will be at $130°$ counterclockwise from the positive y-axis in the yz-plane.

Conclusion: The magnetic force on the segment $bc$ will be at $130°$ counterclockwise from the positive y-axis in the yz-plane.

To determine

The direction of magnetic torque exerted by the force on the wire segment $bc$ with respect to the x-axis.

Answer to Problem 42P

The direction of magnetic torque exerted by the force on the wire segment $bc$ with respect to the x-axis will be along the positive x-axis.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

Since the force on the wire segment $bc$ will rotate the loop counterclockwise with respect to the x-axis, the direction of the torque will be along positive x-axis.

Conclusion: The direction of magnetic torque exerted by the force on the wire segment $bc$ with respect to the x-axis will be along the positive x-axis.

To determine

The direction in which the loop will rotate about the x-axis looking towards the origin along the positive x-axis.

Answer to Problem 42P

The direction of magnetic torque exerted by the force on the wire segment $bc$ with respect to the x-axis will be along the positive x-axis.

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

Only the force on the segment bc will contribute to rotate the loop about the x-axis. Since the force on the wire segment $bc$ will rotate the loop counterclockwise with respect to the x-axis, the loop will rotate counterclockwise about the x-axis..

Conclusion: When looking towards the origin along the positive x-axis, the loop will rotate counterclockwise about the x-axis.

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 42P

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

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

The magnitude of the magnetic moment is defined as,

$\mu =IAN$

• $I$ is the current in the wire
• $N$ is the number of loops in the coil
• $A$ is the area enclosed by the loop

The area of the rectangular loop will be the product of the length and breadth of the loop.

$A=ab$

• $a$ is the length of the loop
• $b$ is the breadth of the loop

Hence the magnetic moment will be,

$\mu =IabN$

Substitute $0.500\text{m}$ for $a$, $0.300\text{m}$ for $b$, $1$ for $N$ and $0.900\text{A}$ for $I$,

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

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

To determine

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

Answer to Problem 42P

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

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

The magnetic moment vector is directed perpendicular to the plane of the loop. Since the loop is in the xy-plane the magnetic moment vector will point parallel to the z-axis. Moreover as the current is flowing clockwise, the magnetic moment will point in negative z-direction. The magnetic field is directed $40°$ with respect to the positive y-axis.

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

$\begin{array}{lll}\theta \hfill & =\hfill & 90°+40°\hfill \\ \hfill & =\hfill & 130°\hfill \end{array}$

Conclusion: The angle between the magnetic moment vector and the magnetic field will be $130°$.

To determine

The torque on the loop.

Answer to Problem 42P

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

Explanation of Solution

Given info: The dimension of the rectangular loop is $0.500\text{m}$ by $0.300\text{m}$. The loop lies on the xy-plane and hinged along the x-axis. The magnitude of the magnetic field is $1.50\text{T}$ and is directed $40°$ with respect to the positive y-axis and lies parallel everywhere to the yz-plane. The current carries by the loop is $0.900\text{A}$.

The torque on the loop is defined as,

$\tau =\mu B\sin \theta$

• $\mu$ is the magnetic moment
• $B$ is the magnetic field
• $\theta$ is the angle between the magnetic field and the magnetic moment

From the previous sections we know the magnitude of the magnetic moment is $0.135\text{A}\cdot {\text{m}}^{\text{2}}$. The angle between the magnetic moment vector and the magnetic field will be $130°$.

Substitute $0.135\text{A}\cdot {\text{m}}^{\text{2}}$ for $\mu$, $130°$ for $\theta$, $1.50\text{T}$ for $B$ to determine the torque,

$\begin{array}{c}\tau =\left(0.135\text{A}\cdot {\text{m}}^{\text{2}}\right)\left(1.50\text{T}\right)\sin 130°\\ =0.155\text{N}\cdot \text{m}\end{array}$

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