**1. The problem statement, all variables and given/known data**

My textbook has two problems, based on the same premises, but the solutions manual solves each of them differently, so I was wondering why? Here are the two problems…

1. In the circuit shown in Fig. P29.49, the capacitor has capacitance C = 20μF and is initially charged to 100V with the polarity shown. The resistor R

_{0}has resistance 10Ω. At time t = 0 the switch is closed. The small circuit is not connected in any way to the large one. The wire of the small circuit has a resistance of 1Ω/m and contains 25 loops. The large circuit is a rectangle 2m by 4m, while the small one has dimensions a = .1m and b = .2m. The distance c is .05m. Both circuits are held stationary. Assume that only the wire nearest the small circuit produces an appreciable magnetic field through it. a)Find the current in the large circuit 200μs after S is closed. b)Find the current in the small circuit 200μs after S is closed.

2. In the circuit in Fig. P29.49, an emf of 90V is added in series with the capacitor and the resistor, and the capacitor is initially uncharged. The emf is placed between the capacitor and the switch, with the positive terminal of the emf adjacent to the capacitor. Otherwise, the two circuits are the same as in the previous problem. The switch is closed at t = 0. When the current in the large circuit is 5A, what are the magnitude and direction of the induced current in the small circuit?

**2. Relevant equations**

∫B dl = Magnetic Flux

i = (V/R)e^{-t/RC}

di/dt = (V/R^{2}C)e^{-t/RC}

B (caused by the single wire of the large circuit) = (μ_{0}I)/(2pi(r))

induced emf = -N d(Flux)/dt

emf = IR

**3. The attempt at a solution**

Okay, so for the first problem…

i = (V/R)e^{-t/RC}

i = (100/10)e^{-(200E-6)/((10)(20E-6))}

i = 3.7A

Then part B…

Flux = ∫B dl

Flux = ∫(μ_{0}I)/(2pi(r)) dr from c to c+a

Flux = ((μ_{0}I)/(2pi))ln(1 + (a/c))

induced emf = -N d(Flux)/dt

induced emf = -N ((μ_{0})/(2pi))ln(1 + (a/c)) **dI/dt
**induced emf = -((25)(4piE-7)ln(3))/(2pi)

**((-3.7)/(200E-6))**

induced emf = .0203

N(perimeter) = total length

25(.6) = 15

R = 1Ω/m

R = 15Ω

emf = IR

.0203 = I(15)

I =.00135A

Then problem 2…

induced emf = -N ((μ_{0})/(2pi))ln(1 + (a/c)) **dI/dt**

i = (V/R)e^{-t/RC}

di/dt = (V/R^{2}C)e^{-t/RC}

di/dt = (1/RC)i

induced emf = -N ((μ_{0})/(2pi))ln(1 + (a/c)) (1/RC)i

induced emf = -((25)(4piE-7)ln(3))/(2pi) (1/((10)(20E-6)) (5)

induced emf = .0275

emf = IR

.0275 = I(15)

I = .00183A

HOWEVER… If you switch methods and attempt to plug the actually derivative into the emf equation in the first problem, the answer is incorrect because RC ≠ 200E-6. And if you use the formula for i to find the amount of time it took to get the current to 5A in the second problem and plug it into the emf equation you get the wrong answer for the same reason. That time doesn’t equal RC. **Why is this and how am I supposed to know which equation to use?**

Thank you!

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