The magnetic field as function of perpendicular distance r from the central axis of cable for 0 < r , r 1 , r 1 < r < r 2 and r > r 2 .

Question

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Answer 1

Question

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

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Answer 2

Question

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

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Answer 3

Question

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

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Answer 4

Question

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

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Answer 5

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

Answer 6

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Answer 7

Chapter 28, Problem 80P

(a)

To determine

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

Answer 8

(a)

Expert Solution

Answer to Problem 80P

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Formula Used:

The relation for the magnetic is given by,

$um=B22μ0$

Calculation:

The magnetic field inside the cylinder $(r<r1)$ and outside the cylinder $(r>r2)$ is zero because the net enclosed current is zero.

$Br<r1=0$

And,

$Br>r2=0$

The magnetic field between the cylinder is calculated as,

$(2πr)B=μ0IB=μ0I2πr$

Conclusion:

Therefore, the magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ is $0$ , for $r1<r<r2$ is $μ0I2πr$ and $r>r2$ is $0$ .

Answer 13

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

Answer 14

(b)

To determine

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

Answer 15

(b)

Expert Solution

Answer to Problem 80P

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated below.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula Used:

The expression for the magnetic energy density in the region between the cylinder is given by,

$um=B22μ0$

The magnetic field between the cylinder is given as,

$B=μ0I2πr$

Calculation:

The magnetic energy density in the region between the cylinder is calculated as,

$um= ( μ 0 I 2πr )22μ0=μ0I28π2r2=12( μ 0 4π)( I 2 π r 2 )um=12( μ 0 / 4π)I2/(π r 2)$

Conclusion:

Therefore, the proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ is stated above.

Answer 20

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

Answer 21

(c)

To determine

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

Answer 22

(c)

Expert Solution

Answer to Problem 80P

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated below.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula Used:

The magnetic energy density in the region between the cylinder is given by,

$um=12(μ04π)(I2πr2)$

The magnetic energy $dUm$ in the cylindrical element of volume $dV$ is given by,

$dUm=umdV=(12( μ 0 4π )( I 2 π r 2 ))(l2πrdr)=μ0I2l4π⋅drr$ ……. (I)

Calculation:

Integrate equation (I) over the limits $r1 to r2$ .

$∫d U m=∫ r 1 r 2 μ 0 I 2 l 4π⋅ drrUm=μ0I2l4π∫ r 1 r 21rdr=μ0I2l4π[lnr]r1r2=μ0I2lln( r 2 / r 1 )4π$

Conclusion:

Therefore, the proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$ is stated above.

Answer 27

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

Answer 28

(d)

To determine

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

Answer 29

(d)

Expert Solution

Answer to Problem 80P

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated below.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Formula Used:

The expression for the total magnetic energy is given by,

$Um=μ0I2lln( r 2/ r 1)4π$ …… (II)

The energy in the magnetic in terms of $L$ and $I$ is given by,

$Um=12LI2$ …….. (III)

Calculation:

The self inductance per unit length is calculated as,

$Um=12LI2L=2( U m )I2=2( μ 0 I 2 lln( r 2 / r 1 ) 4π )I2Ll=μ0ln( r 2 / r 1 )2π$

Conclusion:

Therefore, the proof that self inductance per unit length of the cable arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ is stated above.

Answer 34

Ch. 28 - Prob. 1P Ch. 28 - Prob. 2P Ch. 28 - Prob. 3P Ch. 28 - Prob. 4P Ch. 28 - Prob. 5P Ch. 28 - Prob. 6P Ch. 28 - Prob. 7P Ch. 28 - Prob. 8P Ch. 28 - Prob. 9P Ch. 28 - Prob. 10P

The magnetic field as function of perpendicular distance r from the central axis of cable for 0 < r , r 1 , r 1 < r < r 2 and r > r 2 .

Concept explainers

Videos

The magnetic field as function of perpendicular distance $r$ from the central axis of cable for $0<r, r1$ , $r1<r<r2$ and $r>r2$ .

Answer to Problem 80P

Explanation of Solution

The proof that the magnetic energy density in the region between the cylinder is by $um=12(μ0/4π)I2/(πr2)$ .

Answer to Problem 80P

Explanation of Solution

The proof that the total magnetic energy in a cable of volume of length $l$ is given by $Um=(μ0/4π)I2lln(r2/r1)$

Answer to Problem 80P

Explanation of Solution

The proof that self inductance per unit length of the cab arrangement is given by $L/l=(μ0/2π)ln(r2/r1)$ .

Answer to Problem 80P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The magnetic field as function of perpendicular distance r from the central axis of cable for 0 &lt; r , r 1 , r 1 &lt; r &lt; r 2 and r &gt; r 2 .

Concept explainers

Videos

Answer to Problem 80P

Explanation of Solution

Answer to Problem 80P

Explanation of Solution

Answer to Problem 80P

Explanation of Solution

Answer to Problem 80P

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

The magnetic field as function of perpendicular distance r from the central axis of cable for 0 < r , r 1 , r 1 < r < r 2 and r > r 2 .