Concept explainers
Let đ�œ‡rj = 2 in region 1, defined by 2x + 3y — 4z >1, while pr2 = 5 in region 2 where 2x + 3y - 4z < 1. In region 1, H1 = 50ax - 30ay + 20az A/m. Find (o) HM1; (b) Hr1 (c)Hr2; (d) HN2 (e) 01, the angle between H1 and aN2; (f) 02, the angle between H2 and aN2].
(a)
The value of
Answer to Problem 8.27P
The required value is,
Explanation of Solution
Given Information:
The region 1 is
Calculation:
Both regions are separated by the surface
So, the normal component of magnetic field intensity:
Conclusion:
The required value is,
(b)
The value of
Answer to Problem 8.27P
The required value is,
Explanation of Solution
Given Information:
The region 1 is
Calculation:
The tangential component of magnetic field intensity:
Conclusion:
The required value is,
(c)
The value of
Answer to Problem 8.27P
The required value is,
Explanation of Solution
Given Information:
The region 1 is
Calculation:
The tangential components of magnetic field intensity are continuous across the surface between the regions. So:
:
Conclusion:
The required value is,
(d)
The value of
Answer to Problem 8.27P
The required value is,
Explanation of Solution
Given Information:
The region 1 is
Calculation:
The normal components of magnetic flux density are continuous across the surface between the regions. So,:
Conclusion:
The required value is,
(e)
The angle between
Answer to Problem 8.27P
The angle between
Explanation of Solution
Given Information:
The region 1 is
Calculation:
The unit normal vector:
The required angle:
Conclusion:
The angle between
(f)
The angle between
Answer to Problem 8.27P
The angle between
Explanation of Solution
Given Information:
The region 1 is
Calculation:
The unit normal vector:
The magnetic field intensity:
The required angle,
Conclusion:
The angle between
Want to see more full solutions like this?
Chapter 8 Solutions
Engineering Electromagnetics
- 2, Zs 7 Solve for: Solve for: а.) Z f.)P2 b.) It g.) Q3 c.) St h.) S5 d.) Qt i.) P6 e.) pft j.) Q7 Given: E = 1002 – 90 volts Z4 = 42450 Z1 = 1200 Z5 = 5453.13N Z2 = 2436.87N Z6 = 62 – 90N Z3 = 32 - 30N Z, = 72750arrow_forwardexercisesa Find f'E1? O sketchf). sketch flw)> Center Linearrow_forwardZp Np N₁ Z₁ solve for Zparrow_forward
- Quiz 2-E : Derive a state space model for the system shown in Figure below when the input is fa and the output is z. marrow_forward2, Zs Z7 Solve for: Solve for: h.) S5 i.) P6 j.) Q7 Given: E = 1002 – 90 volts Z4 = 42450 Z1 = 1200 Z5 = 5453.13N Z2 = 2436.87N Z6 = 62 – 90N Z3 = 32 - 30N Z, = 72750arrow_forwardO c. 1.57 sec O d. 0.87 sec O e. 1.94 sec Determine the closed loop poles for a 2nd order underdamped system knowing its settling time (2% criterion) and its natural frequency. ts-5 s and wn-1.0 rad/s Select one: a. s=-0.60 + jo.80 O b.s=-0.50 ± j0.70 O c. s=-0.75 + j1.05 O d. s=-0.80 + j0.60 O e. s=-0.90 ± j1.10 10:33 PM EN 4/19/2021 TOSHIBAarrow_forward
- Since R1=18.10Kohm, R2=67.58Kohm, RC= 1.08Kohm, RE= 3.23Kohm, VCC=14.00V, Beta=241.00 in the circuit given in the figure, calculate the IC current by doing a complete analysis. When performing your operations, 2 steps will be taken after the point. choose the closest one from the stylish ones according to the +/-10% margin of error. There is only 1 correct answer to the question.arrow_forwardRo. 8. For unit- slep responses A,4 find the homafor the system. chowa ahe figure , 20 see tlJae) fimetion ofarrow_forward5a. What is E2=? 5b. What is E3=? 5c. What is the potential difference of b relative to point a (V_ba)?arrow_forward
- GSSecO A current Ia5A flows through an infinitely long regular uwire in emty space. C2 It square perimerer with a side length of distance oỆ de4[cm] from A Second a = 3 [cm] at the current OA +he same plane with this current direction is giver in the figure Lare awailable. Caleulate the enductanee L42 as Cu. %3Darrow_forward2. Consider the system given below. Investigate the stability using describing function. Find nature, amplitude and frequency of limit cycle. S 性 1 $+4 s+2arrow_forward1.11 Given that P - 2a, - a, – 2a. Q = 4a, + 3a, + 2a. R = -a, + a, + 2a. %3D find: (a) P + Q- R|, (b) P-Qx R, (c) Q x P R, (d) (P x Q) (Q x R), (e) (P x Q) x (Q x R), (f) cos 0 pg. (g) sin 0po-arrow_forward
- Introductory Circuit Analysis (13th Edition)Electrical EngineeringISBN:9780133923605Author:Robert L. BoylestadPublisher:PEARSONDelmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage LearningProgrammable Logic ControllersElectrical EngineeringISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education
- Fundamentals of Electric CircuitsElectrical EngineeringISBN:9780078028229Author:Charles K Alexander, Matthew SadikuPublisher:McGraw-Hill EducationElectric Circuits. (11th Edition)Electrical EngineeringISBN:9780134746968Author:James W. Nilsson, Susan RiedelPublisher:PEARSONEngineering ElectromagneticsElectrical EngineeringISBN:9780078028151Author:Hayt, William H. (william Hart), Jr, BUCK, John A.Publisher:Mcgraw-hill Education,