A rectangular loop of wire has dimensions 0.500 m by 0.300 m . The loop is pivoted at the x axis

Name: Problem 26, Chapter 22: Principles of Physics a Calculus Based Text
Uploaded: 2020-11-30T20:21:39-08:00
Duration: 21 min 18 s
Channel: Yaqub Khan

Question

A rectangular loop of wire has dimensions $0.500 \mathrm{m}$ by $0.300 \mathrm{m} .$ The loop is pivoted at the $x$ axis and lies in the $x y$ plane as shown in Figure $P 22.26 .$ A uniform magnetic field of magnitude $1.50 \mathrm{T}$ is directed at an angle of $40.0^{\circ}$ with respect to the $y$ axis with field lines parallel to the $y z$ 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 $a b$ ? (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 $c d ?$ (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 $b c ?$ (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 $a d$ about an axis through the origin? (j) From the point of view of Figure $\mathrm{P} 22.26,$ once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the $x$ axis? $(\mathrm{k})$ Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment vector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (1).

   A rectangular loop of wire has dimensions $0.500 \mathrm{m}$ by $0.300 \mathrm{m} .$ The loop is pivoted at the $x$ axis and lies in the $x y$ plane as shown in Figure $P 22.26 .$ A uniform magnetic field of magnitude $1.50 \mathrm{T}$ is directed at an angle of $40.0^{\circ}$ with respect to the $y$ axis with field lines parallel to the $y z$ 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 $a b$ ?
(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 $c d ?$ (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 $b c ?$ (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 $a d$ about an axis through the origin? (j) From the point of view of Figure $\mathrm{P} 22.26,$ once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the $x$ axis? $(\mathrm{k})$ Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment vector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (1).

Principles of Physics a Calculus Based Text

Raymond A. Serway,… 5th Edition

Chapter 22, Problem 26

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Solved by Expert Yaqub Khan

11/30/2020

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5 \, \mathrm{T}$ at an angle of $40^{\circ}$ with the $y$-axis. We can find the components of the magnetic field as follows: The $y$-component of the magnetic field is $B_y = B \cos(40^{\circ}) = 1.5 \, \mathrm{T} \cos(40^{\circ}) = 1.15 \, \mathrm{T}$. The Show more…

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A rectangular loop of wire has dimensions $0.500 \mathrm{m}$ by $0.300 \mathrm{m} .$ The loop is pivoted at the $x$ axis and lies in the $x y$ plane as shown in Figure $P 22.26 .$ A uniform magnetic field of magnitude $1.50 \mathrm{T}$ is directed at an angle of $40.0^{\circ}$ with respect to the $y$ axis with field lines parallel to the $y z$ 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 $a b$ ? (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 $c d ?$ (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 $b c ?$ (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 $a d$ about an axis through the origin? (j) From the point of view of Figure $\mathrm{P} 22.26,$ once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the $x$ axis? $(\mathrm{k})$ Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment vector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (1).

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Key Concepts

Torque on a Magnetic Dipole

A current loop in a magnetic field experiences a torque that tends to align its magnetic moment with the magnetic field. The torque ? is calculated by ? = ? × B, with its magnitude determined by the sine of the angle between the moment and the field vectors. This torque does not result in a net force when the field is uniform, but it is responsible for the rotational tendency of the loop.

Uniform Magnetic Field Effects

In a uniform magnetic field, while individual segments of a current loop experience forces, these forces often pair up to yield no net translational force, leaving only a torque that causes rotation. Understanding this concept is essential when analyzing how magnetic forces affect the motion of current loops, particularly noting that a uniform field can cause rotation without inducing net linear acceleration.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the force acting on a current-carrying conductor in a magnetic field. When the thumb of the right hand points in the direction of the current and the fingers point in the direction of the magnetic field, the palm indicates the direction of the force. This rule is crucial for visualizing the orientation of forces and torques on different segments of the loop.

Lorentz Force on a Current-Carrying Conductor

This concept describes how a current-carrying conductor experiences a force when placed in a magnetic field. The force F is given by the vector cross product F = I (L × B), where I is the current, L is a vector representing the length and direction of the current, and B is the magnetic field vector. This law underlies the behavior of each segment of the wire in a magnetic field.

Magnetic Moment of a Current Loop

The magnetic moment is a vector quantity that characterizes the strength and orientation of a current loop's magnetic influence. It is defined as the product of the current and the area vector of the loop (? = I·A, with the area vector pointing perpendicular to the loop’s plane according to the right-hand rule). This concept is central to analyzing the torque and energy of the loop in an external magnetic field.

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Transcript

00:01 As we are given that the magnitude of magnetic field is 1 .5 tesla which is directed in yg plan at an angle of 40 degree with y axis.

00:16 So we can write the components of magnetic field as y component of magnetic field equal to b into cosine of 40 degree as magnitude of magnetic field is 1 .5 tesla into cosine of 40 degree this gives us y component of magnetic field equal to 1 .15 tesla similarly we can write the z component of magnetic field as bz equal to b into sine of 40 degree as b is equal to b into sine of 40 degree as b is equal to 1 .5 tesla into sign of 40 degree this gives us the z component of magnetic field equal to 0 .96 tesla now we can write a vector form of magnetic field as b vector equal to y component of magnetic field into unit vector in the direction of y -axis that is j head plus component in g direction into unit vector in g direction this gives us b vector equal to 1 .15 j head plus 0 .96 kh and the unit here is tesla now in the first part of this problem we want to find the direction of magnetic force on segment ab.

02:41 As we have the relation to calculate force on a current carrying conductor in a magnetic field is force equal to current times length of conductor into magnetic field.

03:07 Now we can put different values here.

03:09 The current through segment ab is 0 .90 ampere we can write the length vector of segment ab as length of segment a b is 0 .550 m as the direction of current in segment a b is in the positive wide direction so we have here unit vector is j head cross b vector is 1 .15 j head plus 0 .96 k hat as by using the properties of cross product we have j cross j is equal to 0 and j cross k is equal to i so this gives us the force on segment ab equal to 0 .432 nuten i k so this is the force on segment ab here direction of force on segment ab is along positive x -axis so the force on segment ab tend to rotate this segment in a clockwise direction about the axis so the direction on the dark on segment ab is in the negative z direction now in the c part of this problem we want to find the direction of magnetic force on segment cd again we can write the relation for force as f equal to i times l cross b by plugging different values here the current through segment cd is also 0 .90 ampere length of segment cd is 0 .5 0m0 meter and direction of l is in the direction of current and in segment cd the direction of current is in the negative wire direction that is minus j hat cross b vector is 1 .15 tesla j hat plus 0 .96 tesla k hat.

07:53 Again by using the properties of class product that j cross j is equal to 0 and j cross k is equal to ikf.

08:09 This gives us force on segment cd equal to minus 0 .432 newton i hat.

08:29 This gives us the force on segment cd is in the negative x -axis...

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