PHYSICS FOR SCI & ENGR W WEBASSIGN
PHYSICS FOR SCI & ENGR W WEBASSIGN
10th Edition
ISBN: 9781337888486
Author: SERWAY
Publisher: CENGAGE L
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Chapter 29, Problem 49CP

(a)

To determine

The magnitude and direction of the magnetic field in terms of μ0 , I , r and a at a point P1 .

(a)

Expert Solution
Check Mark

Answer to Problem 49CP

The magnetic field in terms of μ0 , I , r and a at a point P1 is μ0Iπr(2r2a24r2a2) and directed towards the left side.

Explanation of Solution

Given info: The given figure is shown below:

PHYSICS FOR SCI & ENGR W WEBASSIGN, Chapter 29, Problem 49CP

Figure (1)

Write the expression for the current in the solid conductor is,

I=JA

Here,

J is the current density.

I is the current flows in the conductor.

A is the area of the conducting part.

Formula to calculate the net area of the conductor is,

A=π(a2(a2)2(a2)2)=πa22

Replace A by πa22 in equation (1).

I=Jπa22

Rearrange for J .

J=2Iπa2

Write the expression for the area of the solid conducting part is,

As=πa2

Here,

a is the radius of the conducting cylinder.

Write the expression for the area of the cavity part is,

Ac1=π(a2)2=πa24

From the right hand rule, the current is directed out of the page so, the direction of magnetic field at point P1 is towards the left side.

Formula to calculate the magnetic field produces at point P1 considering the conductor as a whole solid cylinder is,

Bs=μ0JAs2πr

Here,

μ0 is the absolute permeability.

Substitute πa2 for As in above equation.

Bs=μ0J(πa2)2πr=μ0Ja22r

Formula to calculate the magnetic field produces at point P1 due to cavity is,

B1=μ0JAc12π(ra2)

Substitute πa24 for Ac1 in above equation.

B1=μ0J(πa24)2π(ra2)=μ0a28(ra2)

Since both the cavities are of same hence the field will be same.

B2=μ0Ja28(r+a2)

Formula to calculate the net magnetic field produces at point P1 is,

B=BsB1B2

Substitute μ0Ja22r for Bs , μ0Ja28(ra2) for B1 and μ0Ja28(r+a2) for B2 .

B=μ0Ja22rμ0Ja28(ra2)μ0Ja28(r+a2)=μ0Ja22(1r14(ra2)14(r+a2))=μ0Ja22(4r2a22r24r(r2a2/4))=μ0Ja22π(2r2a24r(r2a2/4))

Substitute 2Iπa2 for J in above equation.

B=μ0(2Iπa2)a22(2r2a24r(r2a2/4))=μ0Iπr(2r2a24r2a2)

Conclusion:

Therefore, the magnetic field in terms of μ0 , I , r and a at a point P1 is μ0Iπr(2r2a24r2a2) and directed towards the left side.

(b)

To determine

The magnitude and direction of the magnetic field in terms of μ0 , I , r and a at a point P2 .

(b)

Expert Solution
Check Mark

Answer to Problem 49CP

The magnitude of the magnetic field in terms of μ0 , I , r and a at a point P2 is μ0Iπr(2r2+a24r2+a2) and directed towards the top of the page.

Explanation of Solution

From the right hand rule, the current is directed out of the page so, the direction of magnetic field at point P2 is towards the top of the page.

Formula to calculate the magnetic field produces at point P2 considering the conductor as a whole solid cylinder is,

Bs=μ0JAs2πr

Here,

μ0 is the absolute permeability.

Substitute πa2 for As in above equation.

Bs=μ0J(πa2)2πr=μ0Ja22r

Formula to calculate the magnetic field produces at point P2 due to cavity is,

B1=μ0JAc12πr2+(a2)2

Substitute πa24 for Ac1 in above equation.

B1=μ0J(πa24)2πr2+(a2)2=μ0Ja28r2+(a2)2

Since both the cavities are of same hence the field will be same.

B2=μ0Ja28r2+(a2)2

Formula to calculate the net magnetic field produces at point P1 is,

B=BsB1cosθB2cosθ

Substitute μ0Ja22r for Bs , rr2+(a2)2 for cosθ , μ0Ja28r2+(a2)2 for B1 and for B2 .

B=μ0Ja22r(μ0Ja28(ra2))(rr2+(a2)2)(μ0Ja28(r+a2))(rr2+(a2)2)=μ0Ja22r(1r22(r2+a2/4))

Substitute 2Iπa2 for J in above equation.

B=μ0(2Iπa2)a22r(1r22(r2+a2/4))=μ0Iπr(1r22(r2+a2/4))=μ0Iπr(2r2+a24r2+a2)

Conclusion:

Therefore, the magnitude of the magnetic field in terms of μ0 , I , r and a at a point P2 is μ0Iπr(2r2+a24r2+a2) and directed towards the top of the page.

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Chapter 29 Solutions

PHYSICS FOR SCI & ENGR W WEBASSIGN

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