The circuit in the figure below consists of switch S, a 12.0 V ideal battery, a

Question

The circuit in the figure below consists of switch S, a 12.0 V ideal battery, a 17.0 MegaOhms resistor, and an air-filled capacitor. The capacitor has parallel circular plates of radius 4.00 cm, separated by 3.00 mm. At time t = 0, switch S is closed to begin charging the capacitor. The electric field between the plates is uniform. At t = 250 µs, what is the magnitude of the magnetic field within the capacitor, at radial distance 3.00 cm?

The circuit looks like this...(My best attempt)

_________Capacitor_______
|                                       |
Switch                          Resistor
|                                       |
|________E(Battery)______|

Explanation / Answer

Given that Battery voltage,V = 12 V Resistance of the resistor, R = 17*106 Radius of circular plates, r = 4 c m = 0.04 m Seperation between plates, d = 3 mm = 3*10 -3 m ---------------------------------------------------------------------------------------------- capacitance of the capacitor is          C = oA / d                = ( 8.85* 10 -12 ) ( 0.04 2) / ( 3*10-3 )                = 1.4828 *10-11 F Time constant of the circuit is            = RC                = ( 17*106) ( 1.4828*10-11 F )                = 2.52*10 -4 s ------------------------------------------------------------------------------------------ Displacement current              i = ( V / R) (e - t / )               = ( 12 V / 17*106 ) ( e- 2.5*10^-4 s / 2.52*10^ -4 s )                = 2.62* 10 -7 A Therefore magnetic field with in the capacitor is          B = o i r / 2 R 2                   = ( 4*10-7) ( 2.62*10-7A ) ( 0.03) / 2 ( 0.04 ) 2              = 9.825*10-13 T   Given that Battery voltage,V = 12 V Resistance of the resistor, R = 17*106 Radius of circular plates, r = 4 c m = 0.04 m Seperation between plates, d = 3 mm = 3*10 -3 m ---------------------------------------------------------------------------------------------- capacitance of the capacitor is          C = oA / d                = ( 8.85* 10 -12 ) ( 0.04 2) / ( 3*10-3 )                = 1.4828 *10-11 F Time constant of the circuit is            = RC                = ( 17*106) ( 1.4828*10-11 F )                = 2.52*10 -4 s ------------------------------------------------------------------------------------------ Displacement current              i = ( V / R) (e - t / )               = ( 12 V / 17*106 ) ( e- 2.5*10^-4 s / 2.52*10^ -4 s )                = 2.62* 10 -7 A Therefore magnetic field with in the capacitor is          B = o i r / 2 R 2                   = ( 4*10-7) ( 2.62*10-7A ) ( 0.03) / 2 ( 0.04 ) 2              = 9.825*10-13 T   Time constant of the circuit is            = RC                = ( 17*106) ( 1.4828*10-11 F )                = 2.52*10 -4 s ------------------------------------------------------------------------------------------ Displacement current              i = ( V / R) (e - t / )               = ( 12 V / 17*106 ) ( e- 2.5*10^-4 s / 2.52*10^ -4 s )                = 2.62* 10 -7 A Therefore magnetic field with in the capacitor is          B = o i r / 2 R 2                   = ( 4*10-7) ( 2.62*10-7A ) ( 0.03) / 2 ( 0.04 ) 2              = 9.825*10-13 T                = 9.825*10-13 T  

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