Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 32, Problem 24SP

A room has its walls aligned accurately with respect to north, south, east, and west. The north wall has an area of 15 m 2 , the east wall has an area of 12 m 2 , and the floor’s area is 35 m 2 . At the site the Earth’s magnetic field has a value of 0.60 G and is directed 50° below the horizontal and 7.0° east of north. Find the magnetic flux through the north wall, the east wall, and the floor.

Expert Solution & Answer
Check Mark
To determine

The magnetic flux through the north wall, the east wall, and the floor of the room that align accurately with respect to north, south, east, and west if the area of north wall, the east wall, and the floor is 15 m2, 12 m2, and 35 m2, respectively. The Earth’s magnetic field is 0.60G and is directed 50° below the horizontal and 7.0° east of north.

Answer to Problem 24SP

Solution:

0.57 mWb, 56 μWb, and 1.6 mWb.

Explanation of Solution

Given data:

The area of the north wall is 15 m2.

The area of the east wall is 12 m2.

The area of the floor is 35 m2.

The Earth’s magnetic field is 0.60 G.

The direction of the Earth’s magnetic field is 50° below the horizontal line and 7.0° east of north.

Understand that 1 G=104 T.

Formula used:

The expression for flux in terms of magnetic field and area is expressed as

Φ=BA=BAcosθ

Here, B is the magnetic field, A is the area, and θ is the angle between the area normal and magnetic field.

Write the expression for the area of a rectangle:

A=lb

Here, l is the length and b is the breath of the rectangle.

Explanation:

Sketch the approximate diagram of the Earth’s magnetic field and the north wall:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 32, Problem 24SP , additional homework tip  1

In figure (1), θNH is the east–north angle, BE is the Earth’s magnetic field, and θH is the angle between the Earth’s magnetic field and the horizontal line.

Recall the expression of flux through the north wall:

ΦNorth wall=BEANcosθH

Here, AN is the area of the north wall.

Substitute 50° for θH, 15 m2 for AN, and 0.60 G for BE

ΦNorth wall=(0.60 G)(15 m2)cos(50°)=(0.60 G)(15 m2)cos(50°)=(0.60 G(104 T1 G))(15 m2)cos(50°)=5.78×104Wb

The Earth’s magnetic field is directed 7.0° east of north. Hence, the total magnetic flux is calculated as

ΦTotal North wall=(ΦNorth wall)cos7°

Substitute 5.78×104Wb for ΦNorth wall

ΦTotal North wall=(5.78×104Wb)cos7°=0.57 mWb

Hence, the total flux that passes through the north wall is 0.57 mWb.

Sketch the approximate diagram of the Earth’s magnetic field and the east wall:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 32, Problem 24SP , additional homework tip  2

Recall the expression of flux through the east wall:

ΦEast wall=BEAEcosθH

Here, BE is the Earth’s magnetic field, AE is the area of north wall, and θH is the angle between Earth’s magnetic field and the horizontal line.

Substitute 50° for θH, 12 m2 for AE, and 0.60 G for BE

ΦEast wall=(0.60 G)(12 m2)cos(50°)=(0.60 G)(12 m2)cos(50°)=(0.60 G(104 T1 G))(12 m2)cos(50°)=4.63×104Wb

The Earth’s magnetic field is directed 7.0° east of north. Hence, the total magnetic flux that passes through the east wall is calculated as

ΦTotal East wall=(ΦEast wall)sin7°

Substitute 4.63×104Wb for ΦEast wall

ΦTotal East wall=(4.63×104Wb)sin7°=56 μWb

Hence, the total flux that passes through the east wall is 56 μWb.

Sketch the approximate diagram of the Earth’s magnetic field and the floor:

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines), Chapter 32, Problem 24SP , additional homework tip  3

Recall the expression of flux through the floor:

ΦFloor=BEAFcosθH

Here, BE is the Earth’s magnetic field, AF is the area of floor, and θH is the angle between Earth’s magnetic field and the horizontal line.

Understand that the flux through the floor would be vertical component passes through the floor only. Hence, the angle is 90°50°=40°.

Substitute 40° for θH, 35 m2 for AF, and 0.60 G for BE

ΦFloor=(0.60 G)(35 m2)cos(40°)=(0.60 G(104 T1 G))(35 m2)cos(40°)=1.6mWb

Hence, the total flux that passes through the floor is 1.6mWb.

Conclusion:

Therefore, the total flux that passes through the north wall is 0.57 mWb, the total flux that passes through the east wall is 56 μWb, and the total flux that passes through the floor is 1.6mWb.

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