Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 30, Problem 30.47P

A cube of edge length l = 2.50 cm is positioned as shown in Figure P30.47. A uniform magnetic field given by B = (5 i ^ + 4 j ^ + 3 k ^ ) T exists throughout the region. (a) Calculate the magnetic flux through the shaded face. (b) What is the total flux through the six faces?

Chapter 30, Problem 30.47P, A cube of edge length l=2.50 cm is positioned as shown in Figure P30.47. A uniform magnetic field

(a)

Expert Solution
Check Mark
To determine

The magnetic flux through the shaded face.

Answer to Problem 30.47P

The magnetic flux through the shaded face is 3.13mWb .

Explanation of Solution

Given info: The edge length l of the cube is 2.50cm and uniform magnetic field is (5i^+4j^+3k^)T .

Write the expression for the magnetic flux through the shaded face.

ϕ=BAx (1)

Here,

Ax is the area of the square along the x axis.

B is the magnetic field on the shaded portion.

Write the expression for the area of the square along the x axis.

Ax=l2i^

Substitute 2.50cm for l in the above expression for the area of square.

Ax=(2.50cm)2i^=(2.50cm×1m100cm)2i^=6.25×104i^m2

Thus, the area of the square is 6.25×104i^m2 .

Substitute (5i^+4j^+3k^)T for B and 6.25×104i^m2 for Ax in the equation (1) for the magnetic flux through the shaded face.

ϕ=(5i^+4j^+3k^)T6.25×104i^m2=3.125×103T-m2×1Wb1T-m2=3.125×103Wb×1000m-Wb1Wb=3.13m-Wb

Conclusion:

Therefore, the magnetic flux through the shaded face is 3.13mWb .

(b)

Expert Solution
Check Mark
To determine

The total flux through the six faces.

Answer to Problem 30.47P

The total flux through the six faces is 0 .

Explanation of Solution

Given info: The edge length l of the cube is 2.50cm and uniform magnetic field is (5i^+4j^+3k^)T .

Write the expression for the magnetic flux through the six faces.

ϕT=ϕ1+ϕ2+ϕ3+ϕ4+ϕ5+ϕ6 (1)

Here,

ϕ1 is the flux passes through the face 1 .

ϕ2 is the flux passes through the face 2 .

ϕ3 is the flux passes through the face 3 .

ϕ4 is the flux passes through the face 4 .

ϕ5 is the flux passes through the face 5 .

ϕ6 is the flux passes through the face 6 .

Write the expression for the flux passes through the face 1 .

ϕ1=BAx

Here,

Ax is the area of the square along the x axis.

B is the magnetic field on the face 1 .

Write the expression for the flux passes through the face 1 .

ϕ2=BAx

Here,

Ax is the area of the square along the negative x axis.

B is the magnetic field on the face 2 .

Write the expression for the flux passes through the face 3 .

ϕ3=BAy

Here,

Ay is the area of the square along the y axis.

B is the magnetic field on the face 3 .

Write the expression for the flux passes through the face 4 .

ϕ4=BAy

Here,

Ay is the area of the square along the negative y axis.

B is the magnetic field on the face 4 .

Write the expression for the flux passes through the face 5 .

ϕ5=BAz

Here,

Az is the area of the square along the z axis.

B is the magnetic field on the face 5 .

Write the expression for the flux passes through the face 6 .

ϕ6=BAz

Here,

Az is the area of the square along the negative z axis.

B is the magnetic field on the face 6 .

Substitute BAx for ϕ1 , BAx for ϕ2 , BAy for ϕ3 , BAy for ϕ4 , BAz for ϕ5 and BAz for ϕ6 in the above expression for the total magnetic flux passes through the six faces of the cube.

ϕT=BAxBAx+BAyBAy+BAzBAz=0

Conclusion:

Therefore, the total flux through the six faces of the square cube is 0 .

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Chapter 30 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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