Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 34, Problem 6P

(a)

To determine

The electric field the rod creates at the point (x=0,y=20.0cm,z=0) .

(a)

Expert Solution
Check Mark

Answer to Problem 6P

The electric field the rod creates at the point (x=0,y=20.0cm,z=0) is (3.148j^)kV/m .

Explanation of Solution

Given info: The linear density of the rod is 35.0nC/m and the speed is 1.50×107m/s .

The value of permittivity of free space is 8.85×1012F/m .

The figure given below shows the location of the thin rod with respect to axis.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 34, Problem 6P

Figure (1)

The formula for the electric field due to long wire is,

E=λrε0

Here,

ε0 is the permittivity of free space.

λ is the linear density of the electric charge.

r is the distance of the electric field from the origin at y axis.

Substitute 35.0nC/m for λ , 8.85×1012F/m for ε0 and 20.0cm for r in above equation to find E .

E=35.0nC(109C1nC)/m(20.0cm)(102m1cm)(8.85×1012F/m)=(3.148j^)kV/m

Conclusion:

Therefore, the electric field the rod creates at the point (x=0,y=20.0cm,z=0) is (3.148j^)kV/m .

(b)

To determine

The magnetic field the rod creates at the point (x=0,y=20.0cm,z=0) .

(b)

Expert Solution
Check Mark

Answer to Problem 6P

The magnetic field the rod creates at the point (x=0,y=20.0cm,z=0) is (5.25×107k^)T .

Explanation of Solution

Given info: The linear density of the rod is 35.0nC/m and the speed is 1.50×107m/s .

The value of the permeability constant is ×107Tm/A

The expression for the current in the wire is,

I=λv

Here,

v is the speed.

Substitute 35.0nC/m for λ and 1.50×107m/s for v in above equation to find I .

I=(35.0nC(109C1nC)/m)(1.50×107m/s)=0.525A

Thus, the current in the wire is 0.525A .

The formula for the magnetic flux due to wire is,

B=(μ0I2πr)

Here,

μ0 is the permeability constant.

Substitute 20.0cm for r , ×107Tm/A for μ0 and 0.525A for I in above equation to find B .

B=((0.525A)(×107H/m)2π(20.0cm(102m1cm)))=(5.25×107k^)T

Conclusion:

Therefore, the magnetic field the rod creates at the point (x=0,y=20.0cm,z=0) is (5.25×107k^)T .

(c)

To determine

The force exerted on an electron at point (x=0,y=20.0cm,z=0) , moving with velocity (2.40×108)i^m/s .

(c)

Expert Solution
Check Mark

Answer to Problem 6P

The force exerted on an electron at point (x=0,y=20.0cm,z=0) , moving with velocity (2.40×108)i^m/s is 4.83×1016(j^)N .

Explanation of Solution

Given info: The linear density of the rod is 35.0nC/m and the speed is 1.50×107m/s .

The charge on an electron is 1.60×1019C

The Lorentz force on the electron is,

F=qE+qv×B

Here,

q is the charge on an electron.

Substitute 1.60×1019C for q , (3.148×103j^)V/m for E , (2.40×108)i^m/s for v and (5.25×107k^)T for B in above equation to find F .

F=[(1.60×1019C)((3.148×103j^)V/m)+(1.60×1019C)((2.40×108)i^m/s)×((5.25×107k^)T)]=5.04×1016(j^)N+2.06×1017(j^)N=4.83×1016(j^)N

Conclusion:

Therefore, the force exerted on an electron at point (x=0,y=20.0cm,z=0) , moving with velocity (2.40×108)i^m/s is 4.83×1016(j^)N .

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Chapter 34 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

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