Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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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, Vol. 1, 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 .

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Chapter 26 Solutions

Physics for Scientists and Engineers, Vol. 1

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