As shown in the figure(Figure 1) , a 23 micro F capacitor is in series with a re
ID: 1325028 • Letter: A
Question
As shown in the figure(Figure 1) , a 23microF capacitor is in series with a resistance R0 of 10ohm . The capacitor is inititally charged, with the plate near the resistor being positive. Nearby is an isolated small rectangular loop of wire of resistance 1.5 ohm/m and dimensions a = 11.0cmand b = 24.0cm . The distance c to a long straight wire section of the RC circuit is 3.0cm , and the distances to all oher portions of the RC circuit are assumed to be so large that they produce no significant magnetic field in the vicinity of the small rectangular loop. (The figure is not to scale.) Neither the RC circuit nor the small rectangular loop is allowed to move.
At some instant after the switch is closed, the current in the RC circuit is 4.50A . Determine the magnitude of the current in the small rectangular loop at this instant.
Express your answer to two significant figures and include the appropriate units.
As shown in the figure(Figure 1) , a 23microF capacitor is in series with a resistance R0 of 10ohm . The capacitor is inititally charged, with the plate near the resistor being positive. Nearby is an isolated small rectangular loop of wire of resistance 1.5 ohm/m and dimensions a = 11.0cmand b = 24.0cm . The distance c to a long straight wire section of the RC circuit is 3.0cm , and the distances to all oher portions of the RC circuit are assumed to be so large that they produce no significant magnetic field in the vicinity of the small rectangular loop. (The figure is not to scale.) Neither the RC circuit nor the small rectangular loop is allowed to move. At some instant after the switch is closed, the current in the RC circuit is 4.50A . Determine the magnitude of the current in the small rectangular loop at this instant. Express your answer to two significant figures and include the appropriate units.Explanation / Answer
The current is 4.50 A.
Thus, the magnetic field at a location is
B = [uo I] / [2 pi r]
Thus, the total flux in the loop is
B flux = Integral [B dA]
Here, dA = bdr. Thus,
B flux = Integral [[uo I] / [2 pi r] b dr]
Thus,
B flux = { [uo I b]/[2pi] } ln (r2/r1)
Here, r2 = 3.0 cm + 11.0 cm
r1 = 3.0 cm
Then,
B flux = 3.327E-7 T*m^2
Thus, the rate is
B flux/RC = 1.447E-3 V = EMF
The resistance of the small loop is
R = 1.5 ohms/m * 0.70 m = 1.05 ohms
Dividing this by the total resistance,
I = 1.4E-3 A or 1.4 mA. [ANSWER]
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