Lecture Notes: EEEC6430310 Electromagnetic Fields And Waves - Cylindrical Capacitor And Solenoid
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Lecture Notes: EEEC6430310 Electromagnetic Fields And Waves - Cylindrical Capacitor And Solenoid
- 1. EEEC6430310 ELECTROMAGNETIC FIELDS AND WAVES Cylindrical Capacitor And Solenoid FACULTY OF ENGINEERING AND COMPUTER TECHNOLOGY BENG (HONS) IN ELECTRICALAND ELECTRONIC ENGINEERING Ravandran Muttiah BEng (Hons) MSc MIET
- 2. Cylindrical Capacitor Example 1: Design a cylindrical capacitor consisting of two coaxial cylinders of radius 𝑟1 and 𝑟2 meters. With the value of electric flux density on the surface of the imaginary cylinder, 𝐷 = 𝛹 𝐴 , the electric intensity, 𝐸 = 𝐷 𝜀o𝜀r , the capacitance of l meter length must be, 𝐶 = 2π𝜀o𝜀r𝑙 2.3 log10 𝑟2 𝑟1 Farad. The design of a cylindrical capacitor consisting of two coaxial cylinders is shown in figure 1. Now, let us find the value of electric intensity at any point distant 𝑥 meters from the axis of the imaginary coaxial cylinder. Consider an imaginary coaxial cylinder of radius 𝑥 meters and length one meter between the two given cylinders. The electric field between the two cylinders is radial as shown in figure 1. Total flux coming out radially from the curved surface of this imaginary cylinder is 𝑄 coulomb. Area of this curved surface = 2π𝑥 × 1 = 2π𝑥 m2 . Hence, the value of electric flux density on the surface of the imaginary cylinder is, 𝐷 = flux in coulomb area in meter2 = 𝛹 𝐴 = 𝑄 𝐴 Cm−2 1
- 3. 2 ∴ 𝐷 = 𝑄 2π𝑥 Cm−2 The value of electric intensity is, 𝐸 = 𝐷 𝜀o𝜀r or 𝐸 = 𝑄 2π𝜀o𝜀r𝑥 Vm−1 Now, d𝑉 = −𝐸 d𝑥 or 𝑉 = 𝑟2 𝑟1 −𝐸 d𝑥 = 𝑟2 𝑟1 − 𝑄 2π𝜀o𝜀r𝑥 d𝑥 = −𝑄 2π𝜀o𝜀r 𝑟2 𝑟1 1 𝑥 d𝑥
- 4. 3 𝑉 = −𝑄 2π𝜀o𝜀r loge 𝑥 𝑟2 𝑟1 = −𝑄 2π𝜀o𝜀r loge𝑟1 − loge𝑟2 = −𝑄 2π𝜀o𝜀r loge 𝑟1 𝑟2 = 𝑄 2π𝜀o𝜀r loge 𝑟2 𝑟1 𝑄 𝑉 = 2π𝜀o𝜀𝑟 loge 𝑟2 𝑟1 and we know that, loge 𝑟2 𝑟1 = 2.3 log10 𝑟2 𝑟1 ∴ 𝐶 = 2π𝜀o𝜀r 2.3 log10 𝑟2 𝑟1 Fm−1 The capacitance of 𝑙 meter length, 𝐶 = 2π𝜀o𝜀r𝑙 2.3 log10 𝑟2 𝑟1 Farad
- 5. 4 + + + r1 r2 x 1 m Q r2 r1 + + + Q Q Q + + + + + + + + Figure 1: Cylindrical Capacitor. 𝜀r
- 6. 5 Example 2: Design a solenoid having a length 𝑙 and radius of turns 𝑟 be uniformly wound with 𝑁 turns each carrying a current of 𝐼. The winding density number of turns per unit length of the solenoid must be 𝑁 𝑙 . Hence, in a small element of length d𝑥, there must be 𝑁d𝑥 𝑙 turns. A very short element of length of the solenoid can be regarded as a concentrated coil of very short axial length and having 𝑁d𝑥 𝑙 turns. Let d𝐻 be the magnetizing force contributed by the element d𝑥 at any axial point 𝑃 as shown in figure 2. Then, substituting d𝐻 for 𝐻 and 𝑁d𝑥 𝑙 for 𝑁 in 𝐻 = 𝑁𝐼 2𝑟 sin2 𝜃, we get, d𝐻 = 𝑁d𝑥 𝑙 ∙ 𝐼 2𝑟 sin2 𝜃 Solenoid
- 7. 6 Now, d𝑥 ∙ sin 𝜃 = 𝑟 d𝜃 sin 𝜃∗ ∴ d𝑥 = 𝑟 d𝜃 sin2 𝜃 Substituting this value of d𝑥 in the above equation, we get d𝐻 = 𝑁𝐼 2𝑙 sin 𝜃 d𝜃 Total value of the magnetizing force at 𝑃 due to the whole length of the solenoid may be found by integrating the above expression between proper limits. ∴ 𝐻 = 𝑁𝐼 2𝑙 𝜃1 𝜃2 sin 𝜃 d𝜃 = 𝑁𝐼 2𝑙 − cos 𝜃 𝜃1 𝜃2 = 𝑁𝐼 2𝑙 cos 𝜃1 − cos 𝜃2
- 8. 7 The above expression may be used to find the value of 𝐻 at any point of the axis, either inside or outside the solenoid. At the mid-point, 𝜃2 = π − 𝜃1 Hence, cos 𝜃2 = − cos 𝜃1 ∴ 𝐻 = 2𝑁𝐼 2𝑙 cos 𝜃1 = 𝑁𝐼 𝑙 cos 𝜃1 Obviously, when the solenoid is very long, cos 𝜃1 becomes nearly unity. In that case, 𝐻 = 𝑁𝐼 𝑙 ATm−1
- 9. 8 dx 𝜃1 𝜃2 𝜃 d𝜃 r P Figure 2: Solenoid.
- 10. (1) B. L. Theraja and A. K. Theraja, Electrical Technology, S. Chand Publishing, 2005. References 9