Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 26, Problem 52P

(a)

To determine

Whether the magnetic moment of the coil make angle with the unit vector i^ .

(a)

Expert Solution
Check Mark

Answer to Problem 52P

The coil makes an angle of i^ with the y axis.

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  Physics for Scientists and Engineers, Chapter 26, Problem 52P

Figure 1

Calculation:

The angel that the coil makes with the y axis is n^ and the normal is drawn to the coil along the x axis.

In the Figure 1, the angel that the coil makes with y axis is i^ . The normal drawn to the coil have an angle j^ below the positive x-axis.

Conclusion:

Therefore, the coil makes an angle of i^ with the y axis.

(b)

To determine

The expression for n^ in terms of other unit vectors.

(b)

Expert Solution
Check Mark

Answer to Problem 52P

The expression for n^ is 0.8i^0.6j^ .

Explanation of Solution

Given:

The angle θ is 37° .

Formula:

The normal drawn to the coil in the plane of positive x and negative y direction and the expression for n^ is given by,

  n^=nxi^nyj^=cosθi^sinθj^

Calculation:

The value of n^ is calculated as,

  n^=cos(37°)i^sin(37°)j^=0.799i^0.692j^=0.8i^0.6j^

Conclusion:

Therefore, the expression for n^ is 0.8i^0.6j^ .

(c)

To determine

The magnetic moment of the coil.

(c)

Expert Solution
Check Mark

Answer to Problem 52P

The value of the magnetic field is (0.4185Nm)k^ .

Explanation of Solution

Given:

The length of the coil l is 5.00cm

The width of the coil w is 8cm .

The current I in the loop is 1.75A .

The magnetic field density B is 1.5T .

Formula:

The expression for the area of the loop is given by,

  A=lw

The expression to determine the value of μ is given by,

  μ=INAn^

The expression to determine the magnetic moment of the coil is given by,

  τ=μ×B

Calculation:

The area of the loop is calculated as,

  A=lw=(5cm)(8cm)=40cm2

The value of μ is calculated as,

  μ=INAn^=(1.75A)(40 cm2)(50)(0.799i^0.692j^)=(0.279Am2)i^(0.21Am2)j^

The magnetic moment of the coil is calculated as,

  τ=μ×B=[(0.279A m 2)i^(0.21A m 2)j^]×[1.5T]j^=(0.4185Nm)k^

Conclusion:

Therefore, the value of the magnetic moment is (0.4185Nm)k^ .

(d)

To determine

The torque on the coil when the magnetic field is constant.

(d)

Expert Solution
Check Mark

Answer to Problem 52P

The torque on the coil is 0.315J .

Explanation of Solution

Formula:

The expression for the torque on the coil is given by,

  U=μB

Calculation:

The value of the torque on the coil is calculated as,

  U=μB=[(0.279A m 2)i^(0.21A m 2)j^](1.5T)j^=0.315J

Conclusion:

Therefore, the torque on the coil is 0.315J .

(e)

To determine

The potential energy of the coil.

(e)

Expert Solution
Check Mark

Answer to Problem 52P

The potential energy of the coil is 0.315J .

Explanation of Solution

Formula:

The expression to determine the potential energy of the coil is given by,

  U=μB

Calculation:

The potential energy of the coil is calculated as,

  U=μB=[(0.279A m 2)i^(0.21A m 2)j^](1.5T)j^=0.315J

Conclusion:

Therefore, the potential energy of the coil is 0.315J .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
(a) Consider an infinite wire of radius a that carries a current I(t) = Io cos cot When separated into two pieces with a small gap of d <<< a, you would not notice that it is broken when looking from a very far distance. Explain this phenomenon by using the displacement current and electric field in the gap of the broken wire.
Given that the applied field is H = 5 x 104 amp . m-1, calculate the magnetization In Ge ; also the magnetic induction. the value of susceptibility is follow
A cylindrical symmetric and uniform magnetic field has a magnitude given by the following equation: B-[(5)]xsin (2)] What is the magnitude of the electric field (in UV/m) induced in a circular loop of wire with radius p = 0.02 m at t = 3.5 s? The radius R = 0.015 m. x 848.14

Chapter 26 Solutions

Physics for Scientists and Engineers

Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY