A solenoid has N=1568.0 turns, length d=32 cm , and radius b=0.2 cm, (b<<d) . The solenoid is connected via a switch, S1 , to an ideal voltage source with electromotive force ϵ=13 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) Fig. 1
(b) Fig. 2
(a) When a current I=0.260 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)?
unanswered
(c) What is the current (in A) in the circuit a very long time (t>>L/R) after S1 is closed? .
unanswered
(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=151 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?
(a) The magnetic field inside a solenoid can be calculated using the equation:
B = μ₀ * N * I / L
where B is the magnetic field, μ₀ is the permeability of free space (4π × 10^(-7) T m/A), N is the number of turns in the solenoid, I is the current flowing through the solenoid, and L is the length of the solenoid.
Given:
N = 1568.0 turns
I = 0.260 A
L = 32 cm = 0.32 m
Plugging in these values into the equation, we can calculate the magnetic field:
B = (4π × 10^(-7) T m/A) * 1568.0 turns * 0.260 A / 0.32 m
B ≈ 0.00633 T
Therefore, the magnitude of the magnetic field inside the solenoid is approximately 0.00633 Tesla.
(b) The self-inductance of a solenoid can be calculated using the equation:
L = μ₀ * N² * A / d
where L is the self-inductance, μ₀ is the permeability of free space (4π × 10^(-7) T m/A), N is the number of turns in the solenoid, A is the cross-sectional area of the solenoid, and d is the length of the solenoid.
Given:
N = 1568.0 turns
d = 32 cm = 0.32 m
b = 0.2 cm = 0.002 m (radius)
A = π * b² = 3.14 * (0.002 m)²
Plugging in these values into the equation, we can calculate the self-inductance:
L = (4π × 10^(-7) T m/A) * (1568.0 turns)² * 3.14 * (0.002 m)² / 0.32 m
L ≈ 0.0349 H
Therefore, the self-inductance of the solenoid is approximately 0.0349 Henry.
(c) To determine the current in the circuit a very long time after S1 is closed (t >> L/R), we can use the equation:
I(t) = ε/R * (1 - e^(-t/(L/R)))
where I(t) is the current at time t, ε is the electromotive force (voltage) of the source, R is the resistance, L is the self-inductance, and t is the time.
Since t >> L/R, the exponential term e^(-t/(L/R)) approaches zero, and the equation simplifies to:
I(t) = ε/R
Given:
ε = 13 V
R = 25 Ohm
Plugging in these values into the equation, we can calculate the current:
I(t) = 13 V / 25 Ohm
I(t) ≈ 0.52 A
Therefore, the current in the circuit a very long time after S1 is closed is approximately 0.52 Amps.
(d) The energy stored in the magnetic field of the coil a very long time after S1 is closed can be calculated using the equation:
E = (1/2) * L * I²
where E is the energy, L is the self-inductance, and I is the current.
Given:
L = 0.0349 H
I = 0.52 A
Plugging in these values into the equation, we can calculate the energy:
E = (1/2) * 0.0349 H * (0.52 A)²
E ≈ 0.0046 J
Therefore, the energy stored in the magnetic field of the coil a very long time after S1 is closed is approximately 0.0046 Joules.
(a) To find the magnitude of the magnetic field inside the solenoid, we can use the formula for the magnetic field inside a solenoid:
B = μ₀ * N * I / L
where B is the magnetic field, μ₀ is the permeability of free space (4π * 10⁻⁷ Tm/A), N is the number of turns in the solenoid, I is the current flowing through the solenoid, and L is the length of the solenoid.
Given that N = 1568.0 turns and I = 0.260 A, we need to find the length of the solenoid. The length of the solenoid can be calculated using the formula:
L = d - 2r
where d is the total length of the solenoid and r is the radius.
Given that d = 32 cm and b = 0.2 cm, we can calculate:
L = 32 cm - 2 * 0.2 cm = 31.6 cm = 0.316 m
Substituting the values into the formula for the magnetic field, we have:
B = (4π * 10⁻⁷ Tm/A) * 1568.0 * 0.260 A / 0.316 m
Simplifying the expression, we get:
B = 0.0131 T
So, the magnitude of the magnetic field inside the solenoid is 0.0131 Tesla (T).
(b) To find the self-inductance L of the solenoid, we can use the formula for the self-inductance of a solenoid:
L = μ₀ * N² * A / L
where L is the self-inductance, μ₀ is the permeability of free space, N is the number of turns in the solenoid, A is the cross-sectional area of the solenoid, and L is the length of the solenoid.
The cross-sectional area of the solenoid can be calculated using the formula:
A = π * r²
where r is the radius of the solenoid.
Given that b = 0.2 cm, we can calculate the cross-sectional area:
A = π * (0.2 cm)² = π * (0.002 m)²
Substituting the values into the formula for self-inductance, we have:
L = (4π * 10⁻⁷ Tm/A) * (1568.0)² * (π * (0.002 m)²) / 0.316 m
Simplifying the expression, we get:
L = 0.158 H
So, the self-inductance of the solenoid is 0.158 Henry (H).
(c) To find the current in the circuit a very long time after S1 is closed, we can use the formula for the steady-state current in an RL circuit:
I_ss = ϵ / R
where I_ss is the steady-state current, ϵ is the electromotive force, and R is the resistance.
Given that ϵ = 13 V and R = 25 Ohm, we can substitute the values into the formula:
I_ss = 13 V / 25 Ohm
Simplifying the expression, we get:
I_ss = 0.52 A
So, the current in the circuit a very long time after S1 is closed is 0.52 Ampere (A).
(d) To find the energy stored in the magnetic field of the coil a very long time after S1 is closed, we can use the formula for the energy in an inductor:
U = (1/2) * L * I_ss²
where U is the energy stored in the inductor, L is the self-inductance, and I_ss is the steady-state current.
Given that L = 0.158 H and I_ss = 0.52 A, we can substitute the values into the formula:
U = (1/2) * 0.158 H * (0.52 A)²
Simplifying the expression, we get:
U = 0.021 J
So, the energy stored in the magnetic field of the coil a very long time after S1 is closed is 0.021 Joules (J).