A solenoid has N=1760.0 turns, length d=40 cm , and radius b=0.4 cm, (b<<d) . The solenoid is connected via a switch, S1 , to an ideal voltage source with electromotive force ϵ=7 V and a resistor with resistance R=25 Ohm . Assume all the self-inductance in the circuit is due to the solenoid. At time t=0 , S1 is closed while S2 remains open.

(a) When a current I=0.112 A is flowing through the outer loop of the circuit (i.e. S1 is still closed and S2 is still open), what is the magnitude of the magnetic field inside the solenoid (in Tesla)?
(b) What is the self-inductance L of the solenoid (in H)?
(c) What is the current (in A) in the circuit a very long time (t>>L/R) after S1 is closed? .
(d) How much energy (in J) is stored in the magnetic field of the coil a very long time (t>>L/R) after S1 is closed?

For the next part, assume that a very long time (t>>L/R) after the switch S1 was closed, the voltage source is disconnected from the circuit by opening switch S1. Simultaneously, the solenoid is connected to a capacitor of capacitance C=751 μF by closing switch S2. Assume there is negligible resistance in this new circuit.
(e) What is the maximum amount of charge (in Coulombs) that will appear on the capacitor?
(f) How long does it take (in s) after S1 is opened and S2 is closed before the capacitor first reaches its maximum charge?

To solve these questions, we will use various formulas and concepts related to circuits and electromagnetic induction. Let's go through each question step-by-step.

(a) To find the magnetic field inside the solenoid, we'll use the formula for the magnetic field inside a solenoid:

B = (μ₀ * N * I) / l

where B is the magnetic field, μ₀ is the magnetic constant (4π x 10^-7 T m/A), N is the number of turns, I is the current, and l is the solenoid length.

Plugging in the given values:
N = 1760 turns
I = 0.112 A
l = 0.4 m

B = (4π x 10^-7 T m/A) * (1760 turns) * (0.112 A) / (0.4 m)
B ≈ 5 × 10^-4 T

Therefore, the magnitude of the magnetic field inside the solenoid is approximately 5 × 10^-4 Tesla.

(b) The self-inductance L of a solenoid is given by the formula:

L = (μ₀ * N² * A) / l

where A is the cross-sectional area of the solenoid (π * b²) and other variables have their previous meanings.

Plugging in the given values:
N = 1760 turns
b = 0.4 cm = 0.004 m
l = 0.4 m

A = π * (0.004 m)² = 1.6 x 10^-5 m²

L = (4π x 10^-7 T m/A) * (1760 turns)² * (1.6 x 10^-5 m²) / (0.4 m)
L ≈ 0.014 H

Therefore, the self-inductance of the solenoid is approximately 0.014 Henry.

(c) For t >> L/R (very long time), the current in the circuit will reach a steady state. In a steady state, the voltage across a resistor in a series circuit is given by Ohm's Law:

V = I * R

The electromotive force (ε) of the voltage source is equal to the potential difference across the inductor (L) and resistor (R) in a steady state:

ε = L * (dI/dt) + I * R

At very long time, the derivative dI/dt is nearly zero, representing a steady state with constant current. This equation simplifies to:

ε = I * R

Substituting the given values:
ε = 7 V
R = 25 Ω

7 V = I * 25 Ω
I ≈ 0.28 A

Therefore, the current in the circuit a very long time after S1 is closed is approximately 0.28 Ampere.

(d) The energy stored in the magnetic field of the solenoid is given by the formula:

U = (1/2) * L * I²

Substituting the given values:
L = 0.014 H
I = 0.28 A

U = (1/2) * 0.014 H * (0.28 A)²
U ≈ 5.4 mJ (millijoules)

Therefore, the energy stored in the magnetic field of the solenoid a very long time after S1 is closed is approximately 5.4 millijoules.

(e) When the solenoid is connected to the capacitor, the energy stored initially in the magnetic field of the solenoid (from part (d)) gets transferred to the electric field of the capacitor. The maximum charge (Q) on a capacitor is given by the formula:

Q = C * V

where C is the capacitance of the capacitor and V is the voltage across the capacitor.

Substituting the given values:
C = 751 μF = (751 x 10^-6 F)
V = ε = 7 V

Q = (751 x 10^-6 F) * (7 V)
Q ≈ 5.26 x 10^-3 C (Coulombs)

Therefore, the maximum amount of charge on the capacitor is approximately 5.26 x 10^-3 Coulombs.

(f) The time it takes for the capacitor to reach its maximum charge can be calculated using the formula for the charging of a capacitor through a resistor:

t = R * C

Substituting the given values:
R = 25 Ω
C = 751 μF = (751 x 10^-6 F)

t = (25 Ω) * (751 x 10^-6 F)
t ≈ 1.88 x 10^-4 s (seconds)

Therefore, it takes approximately 1.88 x 10^-4 seconds for the capacitor to reach its maximum charge after S1 is opened and S2 is closed.

Note: In both parts (e) and (f), it is assumed that the initial energy stored in the magnetic field of the solenoid is fully transferred to the capacitor, neglecting any energy losses.