1) An inductor with an inductance of L and a resistance of R is connected to the

Question

1) An inductor with an inductance of L and a resistance of R is connected to the terminals of a battery with an emf of E and negligible internal resistance.

a) Find the initial rate of increase of current in the circuit.

b) Find the rate of increase of current at the instant when the current is I.

c) Find the current at a time t after the circuit is closed.

d) Find the final steady-state current.

2) A coil containing N = 430 turns with radius r = 3.80 cm , is placed in a uniform magnetic field that varies with time according to B=(C1t+C2t4), where C1 = 1.30×102 T/s and C2 = 2.75×105 T/s4 . The coil is connected to a resistor of resistance R = 700 , and its plane is perpendicular to the magnetic field. The resistance of the coil can be neglected.

a) Find the magnitude of the induced emf E in the coil as a function of time.

Write your answer as an expression in terms of some or all of the variables given in the problem (C1, C2, N, R, and r) and any necessary constants.

b) What is the current I in the resistor at time t0 = 5.40 s ?   Enter your answer in amperes.

Explanation / Answer

1) given inductance = L, resistance = R, emf of battery = E
so, this can be assumed to be a series RL circuit
so in a series RL circuit at time t
E = -Ldi/dt + iR
Ldi/dt = iR - E
Ldi/(i - E/R) = Rdt
integrating
L*ln(i - E/R) = Rt + c
now at t = 0, the inductor acts as open circuit
so, at t= 0, i = 0
L*ln(-E/R) = c
hence L(ln(1 - iR/E)) = Rt
i = E/R
so initial rate of change of current = di/dt = (iR-E)/L = 0
2) when current is i
di/dt = (iR - E)/L
3) L*ln(1 - iR/E) = Rt
i = E(1 - e^[(Rt)/L])/R
4) steady state current is when inductor starts to behave as close circuit
so, i = E/R

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