view-notes-pcs110-009

# view-notes-pcs110-009 - CHAPTER 34 MAXWELLS EQUATIONS AND...

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CHAPTER 34 MAXWELL’S EQUATIONS AND ELECTROMAGNETIC WAVES Section 34-2: Ambiguity in Ampère’s Law Problem 1. A uniform electric field is increasing at the rate of 15 . V/m s. m What is the displacement current through an area of 10 . cm 2 at right angles to the field? Solution Maxwell’s displacement current is e f m 0 12 885 10 15 1 133 ∂ = × = - E t = ( . )( . )( ) . F/m V/m s cm nA. 2 (See Equations 34-1 and 24-2.) Problem 2. A parallel-plate 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 f e f e E D E EA V d A I t A d dV dt = = = ∂ = = ( ) , ( ) = = = = and 0 0 ( . )( ( ) ( . ) . 8 85 10 10 220 0 5 389 12 × = - F/m cm) V/ms cm A. 2 = m Problem 3. A parallel-plate 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 = e f 0 = . For a parallel-plate capacitor, E q A I EA t D = = ∂ = = = e e 0 0 , ( ) so e e 0 0 ∂ = ( ) 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 = m e 0 0 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 34-1, if we take a magnetic field line of radius r R , the electric flux and displacement current become f p e p E D R V d I R d dV dt = = 2 0 2 = = = , ( )( ). and Then Ampère’s law gives B I r R rd dV dt D = = m p m e 0 0 0 2 2 2 = = = ( )( ). Problem 5. A parallel-plate 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?

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802 CHAPTER 34 Solution (a) As explained in Example 34-1, cylindrical symmetry and Gauss’s law for magnetism require that the B -field lines be circles around the symmetry axis, as in Fig. 34-5. For a radius, r , less than the radius of the plates, R , the displacement current is I d dt d dt dA r dE dt D E = = z = e f e e p 0 0 0 2 = = = ( ) ( ), E where the integral is over a disk of radius r centered between the plates. Maxwell’s form of Ampère’s law gives H B = = d rB I D l 2 0 p m , where the line integral is around the circumference of the disk. Thus, B r dE dt r dE dt c c = = 1 2 0 0 2 2 m e ( ) ( ) , = = = where is the speed of light (Equation 34-16). On the symmetry axis, r B = = 0 0 , . so (b) For r R B = < = × = × - 15 015 10 3 10 8 33 10 1 2 6 8 2 13 cm m V/m s m/s T , ( . )( ) ( ) . .
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view-notes-pcs110-009 - CHAPTER 34 MAXWELLS EQUATIONS AND...

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