induced emf = A dB/dt = 1*2*5 =10 V time constant T = RC = 20 *90*10^-6 = 1.80 m
ID: 1774944 • Letter: I
Question
induced emf = A dB/dt = 1*2*5 =10 V
time constant T = RC = 20 *90*10^-6 = 1.80 ms
Current = 10/20= 0.5A anticlockwise
b.) How long will it take the capacitor to charge to 25% of its final charge?
______________ms
and which plate will be positively charged? Top, Bottom
c.) How much charge will the capacitor hold after a long time (assuming this magnetic field rate increase is maintained)?
Q(t)=Qf(1-e^t/rc)
using Q=VC
V=IR so,
Q=IRC=(0.5)(20)(9.0)*10^-6
Qf=9*10^-6
final charge = CV = 90 *10^-6 *10 = 0.9mC
Why does the magnetic field need to be changing? Why can't a very strong but constant magnetic field charge the capacitor?
Explanation / Answer
(A) induced emf = d(flux)/dt
= A dB/dt
= (1 x 2)(5) = 10 Volt
T = R C = 1.80 x 10^-3 = 1.8 ms
I = 10 / 20 = 0.5 A
(B) V = V0[1 - e^(-t/T)]
0.25 = 1 - e^(-t/1.8)
t = 0.52 ms ......Ans
bottom
(C) after long time V = 10 Volt
Q = C V = 900 uC
(emf induced only when flux change. if it is zero then e = 0 )
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