Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9780078028151
Author: Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher: Mcgraw-hill Education,
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Chapter 8, Problem 8.33P
To determine

(a)

The expression of Hϕ(ρ) for z>0.

Expert Solution
Check Mark

Answer to Problem 8.33P

The required expression is, Hϕ(ρ)=23.9ρA/m.

Explanation of Solution

Given Information:

The toroid is having square cross section, 2.5cm<ρ<3.5cm , 0.5cm<z<0.5cm . The flux in aϕ direction is established by an mmf of 150At.

   μr={200.5cm<z<0100<z<0.5cm

Calculation:

The magnetic field intensity,

   Hϕ(ρ)=NI2πρ=1502πρ =23.9ρA/m

Conclusion:

The required expression is, Hϕ(ρ)=23.9ρA/m.

To determine

(b)

The expression of Bϕ(ρ) for z>0.

Expert Solution
Check Mark

Answer to Problem 8.33P

The required expression is, Bϕ(ρ)=3×104ρWb/m2.

Explanation of Solution

Given Information:

The toroid is having square cross section, 2.5cm<ρ<3.5cm , 0.5cm<z<0.5cm . The flux in aϕ direction is established by an mmf of 150At.

   μr={ 20 0.5cm<z<0 10 0<z<0.5cmHϕ(ρ)=23.9ρA/m

Calculation:

The magnetic flux density,

   Bϕ(ρ)=μ0μrHϕ=(4π× 10 7)(10)23.9ρ =3× 10 4ρWb/m2

Conclusion:

The required expression is, Bϕ(ρ)=3×104ρWb/m2.

To determine

(c)

The expression of ϕz>0.

Expert Solution
Check Mark

Answer to Problem 8.33P

The required value is, ϕz>0=5×107Wb.

Explanation of Solution

Given Information:

The toroid is having square cross section, 2.5cm<ρ<3.5cm , 0.5cm<z<0.5cm . The flux in aϕ direction is established by an mmf of 150At.

   μr={ 20 0.5cm<z<0 10 0<z<0.5cmBϕ(ρ)=3× 10 4ρWb/m2

Calculation:

The magnetic flux,

   ϕz>0=SBdS=00.5× 10 2 2.5× 10 2 3.5× 10 2 3× 10 4 ρ dρdz =3×104(0.5×102)ln(3.5× 10 22.5× 10 2) =5×107Wb

Conclusion:

The required value is, ϕz>0=5×107Wb.

To determine

(d)

The expressions of Hϕ(ρ) , Bϕ(ρ) , and ϕz<0 for z<0.

Expert Solution
Check Mark

Answer to Problem 8.33P

The required expressions are,

   Hϕ(ρ)=23.9ρA/mBϕ(ρ)=6× 10 4ρWb/m2ϕz<0=1.0×106Wb.

Explanation of Solution

Given Information:

The toroid is having square cross section, 2.5cm<ρ<3.5cm , 0.5cm<z<0.5cm . The flux in aϕ direction is established by an mmf of 150At.

   μr={200.5cm<z<0100<z<0.5cm

Calculation:

The magnetic field intensity,

   Hϕ(ρ)=NI2πρ=1502πρ =23.9ρA/m

The magnetic flux density,

   Bϕ(ρ)=μ0μrHϕ=(4π× 10 7)(20)23.9ρ =6× 10 4ρWb/m2

The magnetic flux,

   ϕz<0=SBdS=0.5× 10 20 2.5× 10 2 3.5× 10 2 6× 10 4 ρ dρdz =6×104(0.5×102)ln(3.5× 10 22.5× 10 2) =1.0×106Wb

Conclusion:

The required expressionsare,

   Hϕ(ρ)=23.9ρA/mBϕ(ρ)=6× 10 4ρWb/m2ϕz<0=1.0×106Wb.

To determine

(e)

The value of ϕtotal.

Expert Solution
Check Mark

Answer to Problem 8.33P

The required value is, ϕtotal=1.5×106Wb.

Explanation of Solution

Given Information:

The toroid is having square cross section, 2.5cm<ρ<3.5cm , 0.5cm<z<0.5cm . The flux in aϕ direction is established by an mmf of 150At.

   μr={ 20 0.5cm<z<0 10 0<z<0.5cmϕz<0=1.0×106Wbϕz>0=5×107Wb

Calculation:

The totalflux,

   ϕtotal=ϕz<0+ϕz>0=1.0×106+5.0×107 =1.5×106Wb

Conclusion:

The required value is, ϕtotal=1.5×106Wb.

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