This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 32 p. 1 CHAPTER 32 – Maxwell’s Equations and Electromagnetic Waves 1. The electric field between the plates depends on the voltage: E = V / d , so d E /d t = (1/ d ) d V /d t = (1/1.3 × 10 –3 m)(120 V/s) = 9.2 × 10 4 V/m · s . 2. The displacement current is I D = Å A (d E /d t ) = (8.85 × 10 –12 C 2 /N · m 2 )(0.038 m) 2 (2.0 × 10 6 V/m · s) = 2.6 × 10 –8 A . 3. The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from I D = Å A (d E /d t ); 1.8 A = (8.85 × 10 –12 C 2 /N · m 2 )(0.0160 m) 2 d E /d t , which gives d E /d t = 7.9 × 10 14 V/m · s . 4. The current in the wires is the rate at which charge is accumulating on the plates and must also be the displacement current in the capacitor. Because the location is outside the capacitor, we can use the expression for the magnetic field of a long wire: B = ( µ /4p)2 I D / R = (10 –7 T · m/A)2(35.0 × 10 –3 A)/(0.100 m) = 7.00 × 10 –8 T . After the capacitor is fully charged, all currents will be zero, so the magnetic field will be zero . 5. The electric field between the plates depends on the voltage: E = V / d , so d E /d t = (1/ d ) d V /d t . Thus the displacement current is I D = Å A (d E /d t ) = ( Å A / d )(d V /d t ) = C d V /d t . 6. ( a ) The current in the wires is the rate at which charge is accumulating on the plates and must also be the displacement current in the capacitor: I max = I D,max = 35 µ A . ( b ) At any instant, the charge on the plates is Q = å C , so the current is I = d Q /d t = C d å /d t = – ω å C sin ω t , or I max = ω å Å p R 2 / d ; 3 5 × 10 –6 A = 2p(96.0 Hz) å (8.85 × 10 –12 C 2 /N · m 2 )p(0.025 m) 2 /(2.0 × 10 –3 m), which gives å = 6.7 × 10 3 V . ( c ) We can find the maximum value of d Φ E /d t from the maximum value of the displacement current: I D,max = Å (d Φ E /d t ) max ; 3 5 × 10 –6 A = (8.85 × 10 –12 C 2 /N · m 2 )(d Φ E /d t ) max , which gives (d Φ E /d t ) max = 4.0 × 10 6 V · m/s . 7. Gauss’s law for electricity and Ampere’s law will not change. From the analogy to Gauss’s law for electric fields, where Q is the source, Q m would be the source of the magnetic field, so we have ı B · d A = µ Q m . From the analogy to Ampere’s law, we have an additional “current” contribution to Faraday‘s law. ı E · d ¬ = µ d Q m /d t – d Φ B /d t . The d Q m /d t term corresponds to an electric field created by the “current” of magnetic monopoles. 8. The electric field is E = cB = (3.00 × 10 8 m/s)(17.5 × 10 –9 T) = 5.25 V/m . Chapter 32 p. 2 9. We find the magnetic field from E = cB ; 0.43 × 10 –4 V/m = (3.00 × 10 8 m/s) B , which gives B = 1.4 × 10 –13 T ....
View
Full Document
 Spring '09
 Fielding
 Physics, Current, Magnetic Field, m/s, Ampere

Click to edit the document details