The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Question

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Answer 1

Question

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C

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Answer 2

Question

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C

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Answer 3

Question

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C

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Answer 4

Question

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C

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Answer 5

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C

Answer 6

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C

Answer 7

Chapter 22, Problem 42QAP

To determine

The electric field from time intervals 0 to 4 ms, 4 to 10 ms and 10 to 14 ms.

Answer 8

Expert Solution & Answer

Answer to Problem 42QAP

The mass of the mediating particle. is, E0→4=−6.62×10−4N/C, E4→10=4.4×10−6N/C and E10→14=0N/C

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Diameter of ring, d=1.5cm

Radius of the ring, r=0.75cm=0.75×10−2m

Formula used:

Electric field is given by,

E=ϕΔt

Where,

E =Electric field

ϕ = Electric flux

Δt =Time interval

R =Radius of earth

Electric field form t=0 to t=4ms is given by,

E0→4=ΔϕΔtBut, Δϕ=(BA)t=0−(BA)t=4=0−(BA)t=4∴E0→4=− ( BA ) t=4ΔtE0→4=−B( π r 2 )ΔtE0→4=−15× 10 −3( 3.14× (0.75× 10 −2 ) 2 )4× 10 −3E0→4=−6.62×10−4N/C

Electric field form t=4ms to t=10ms is given by,

E4→10=ΔϕΔtBut, Δϕ=(BA)t=4−(BA)t=10∴E4→10= ( BA ) t=4− ( BA ) t=10ΔtE4→10=( π r 2 )( B t=4 − B t=10 )ΔtE4→10=( 3.14× (0.75× 10 −2 ) 2 )( 15× 10 −3 −( −10×10−3 ))6×× 10 −3E4→10=4.4×10−6N/C

Electric field form t=10ms to t=14ms is given by,

E10→14=ΔϕΔtBut, Δϕ=(BA)t=10−(BA)t=14∴E10→14= ( BA ) t=10− ( BA ) t=14ΔtE10→14=( π r 2 )( B t=10 − B t=14 )ΔtE10→14=( 3.14× (0.75× 10 −2 ) 2 )( −10× 10 −3 −( −10×10−3 ))6×× 10 −3E10→14=0N/C