P32_041 - between the plates I d = ( r 2 /R 2 ) i d , where...

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41. (a) At any instant the displacement current i d in the gap between the plates equals the conduction current i in the wires. Thus i max = i d max =7 . 60 µ A. (b) Since i d = ε 0 ( d Φ E /dt ), µ d Φ E dt max = i d max ε 0 = 7 . 60 × 10 6 A 8 . 85 × 10 12 F / m =8 . 59 × 10 5 V · m / s . (c) According to problem 29, i d = C dV dt = ε 0 A d dV dt . Now the potential diference across the capacitor is the same in magnitude as the em± o± the generator, so V = E m sin ωt and dV/dt = ω E m cos ωt .Thu s , i d = ε 0 E m d cos ωt and i d max = ε 0 E m d . This means d = ε 0 E m i d max = (8 . 85 × 10 12 F / m) π (0 . 180m) 2 (130 rad / s)(220V) 7 . 60 × 10 6 A =3 . 39 × 10 3 m , where A = πR 2 was used. (d) We use the Ampere-Maxwell law in the ±orm H ~ B · d~s = µ 0 I d , where the path o± integration is a circle o± radius r between the plates and parallel to them. I d is the displacement current through
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Unformatted text preview: between the plates I d = ( r 2 /R 2 ) i d , where i d is the total displacement current between the plates and R is the plate radius. The ²eld lines are circles centered on the axis o± the plates, so ~ B is parallel to d~s . The ²eld has constant magnitude around the circular path, so H ~ B · d~s = 2 πrB . Thus, 2 πrB = µ µ r 2 R 2 ¶ i d and B = µ i d r 2 πR 2 . The maximum magnetic ²eld is given by B max = µ i d max r 2 πR 2 = (4 π × 10 − 7 T · m / A)(7 . 6 × 10 − 6 A)(0 . 110m) 2 π (0 . 180m) 2 = 5 . 16 × 10 − 12 T ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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