solutions_38 - PHY2061 R. D. Field Chapter 38 Solutions...

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PHY2061 R. D. Field Solutions Chapter 38 Page 1 of 10 C R ε Chapter 38 Solutions Problem 1: A parallel-plate capacitor whose plates have radius R is being charged by a current I . What is the magnitude of the magnetic field ( in Tesla ) between the plates a distance r = R/2 from the center? Answer: kI/R Solution: We know from Ampere’s Law (with J = 0 inside the capacitor) that r r Bd l d dt E ⋅= µε 00 Φ . If I choose a loop of radius r within the capacitor then Φ E = E π r 2 and = = 2 0 2 0 0 2 0 0 ) ( 2 R I r dt dE r r rB π ε µ , where I have used E = Q/( ε 0 A) , A = π R 2 , and I = dQ/dt . Solving for B gives 2 2 ) ( R kIr r B = , which for r = R/2 becomes B = kI/R. Problem 2: A parallel-plate capacitor whose plates have radius r is being charged through a resistor R by an EMF ε as shown in the Figure . If the switch is closed at t = 0 , what is the maximum magnitude of the magnetic field between the plates of the capacitor a distance r/2 from the center? (Note: k = µ 0 /4 π ) Answer: k ε /(rR) Solution: We know from Ampere’s Law (with J = 0 inside the capacitor) that r r l d E Φ .
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PHY2061 R. D. Field Solutions Chapter 38 Page 2 of 10 If I choose a loop of radius r a within the capacitor then Φ E = E π r a 2 and = = 2 0 2 0 0 2 0 0 ) ( 2 r I r dt dE r r B r a a a a π ε µ , where I have used E = Q/( ε 0 A) , A = π r 2 , and I = dQ/dt . The maximum magnetic field occurs when the current is maximum and solving for B max at r a = r/2 gives rR k r kI B = = max max , where I used I max = ε /R (which occurs at t = 0). Problem 3: The potential difference, V C , across a 3 Farad parallel-plate capacitor whose plates have radius R = 2 cm varies with time and is given by V C (t) = at , with a = 10 V/s . What is the magnitude of the magnetic field ( in microT ) between the plates of the capacitor a distance R from the center? Answer: 300 Solution: We know from Ampere’s Law ( with J = 0 inside the capacitor ) that r r Bd l d dt E ⋅= µε 00 Φ . If I choose a loop of radius R within the capacitor then Φ E = E π R 2 and = = 2 0 2 0 0 2 0 0 ) ( 2 R I R dt dE R R RB , where I have used E = Q/( ε 0 A) , A = π R 2 , and I = dQ/dt . Solving for B gives V m s V F A Tm dt dV C R k dt dQ R k R B 300 10 2 ) / 10 )( 3 )( / 10 ( 2 2 2 ) ( 2 7 = × = = = where I used Q = CV which implies that dQ/dt = C dV/dt with dV/dt = a .
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PHY2061 R. D. Field Solutions Chapter 38 Page 3 of 10 Problem 4: A 2 milliFarad parallel-plate capacitor whose plates are circular with radius R = 0.5 cm is connected to an EMF-source that increases uniformly with time at a rate of 150 V/s . What is the magnitude of the magnetic field (in microTesla ) between the plates of the capacitor a distance r = R from the center? Answer: 12 Solution: We know from Ampere’s Law (with J = 0 inside the capacitor) that r r Bd l d dt E ⋅= µε 00 Φ .
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This note was uploaded on 05/31/2011 for the course PHY 2061 taught by Professor Fry during the Spring '08 term at University of Florida.

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solutions_38 - PHY2061 R. D. Field Chapter 38 Solutions...

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