This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CHAPTER 34MAXWELLS EQUATIONS AND ELECTROMAGNETIC WAVES Section 342:Ambiguity in Ampres Law Problem 1. A uniform electric field is increasing at the rate of 15 . V/m s. What is the displacement current through an area of 10 . cm 2 at right angles to the field? Solution Maxwells displacement current is 12 8 85 10 15 1 133 = = E t = ( . )( . )( ) . F/m V/m s cm nA. 2 (See Equations 341 and 242.) Problem 2. A parallelplate capacitor has square plates 10 cm on a side and 0 50 . cm apart. If the voltage across the plates is increasing at the rate of 220 V/ms, what is the displacement current in the capacitor? Solution The electric field is approximately uniform in the capacitor, so E D E EA V d A I t A d dV dt = = = = = ( ) , ( ) = = = = and ( . )( ( ) ( . ) . 8 85 10 10 220 0 5 389 12 = F/m cm) V/ms cm A. 2 = Problem 3. A parallelplate capacitor of plate area A and spacing d is charging at the rate dV dt = . Show that the displacement current in the capacitor is equal to the conduction current flowing in the wires feeding the capacitor. Solution The displacement current is I t D E = = . For a parallelplate capacitor, E q A I EA t D = = = = = , ( ) so = ( ) q t dq dt = = = . But dq dt = is just the conduction current (the rate at which charge is flowing onto the capacitor plates); hence I I D = . Problem 4. A capacitor with circular plates is fed with long, straight wires along the axis of the plates. Show that the magnetic field outside the capacitor, in a plane that passes through the interior of the capacitor and is perpendicular to the axis, is given by B R rd dV dt = 2 2 Here R is the plate radius, d the spacing, dV dt = the rate of change of the capacitor voltage, and r the distance from the axis. Solution In Exercise 341, if we take a magnetic field line of radius r R , the electric flux and displacement current become E D R V d I R d dV dt = = 2 2 = = = , ( )( ). and Then Ampres law gives B I r R rd dV dt D = = 2 2 2 = = = ( )( ). Problem 5. A parallelplate capacitor has circular plates with radius 50 cm and spacing 1.0 mm. A uniform electric field between the plates is changing at the rate 10 . MV/m s. What is the magnetic field between the plates (a) on the symmetry axis, (b) 15 cm from the axis, and (c) 150 cm from the axis? Solution (a) As explained in Example 341, cylindrical symmetry and Gausss law for magnetism require that the Bfield lines be circles around the symmetry axis, as in Fig. 345. For a radius, r , less than the radius of the plates, R , the displacement 31 CHAPTER 34 current is I d dt d dt dA r dE dt D E = = z = 2 = = = ( ) ( ), E where the integral is over a disk of radius r centered between the plates. Maxwells form of Ampres law gives H B = = d rB I D 2 , where the line integral is around the circumference of the disk. Thus, circumference of the disk....
View Full
Document
 Spring '08
 Hoston,Amahd,Bakanowski
 Current

Click to edit the document details