The inductor in the circuit has a resistance of 100 ohms connected to a long solenoid with a loop density of 100 turns per centimeter and 59 H inductance connected to a 12 V battery.
When the switch has been closed for a long time, what is the magnetic field magnitude inside the inductor in Tesla?
To find the magnetic field magnitude inside the inductor, we can use the formula for the magnetic field inside a solenoid:
B = μ₀ * N * I / L
where:
B is the magnetic field magnitude,
μ₀ is the permeability of free space (4π × 10⁻⁷ T m/A),
N is the number of turns per unit length (in this case, 100 turns/cm = 100 turns / 0.01 m = 10,000 turns/m),
I is the current flowing through the inductor,
L is the length of the solenoid.
In this case, the inductor has an inductance of 59 H and is connected to a 12 V battery. When the switch has been closed for a long time, the inductor will reach a steady-state condition, and the current flowing through the inductor can be calculated using Ohm's Law:
I = V / R
where V is the voltage (12 V) and R is the resistance (100 Ω) of the inductor.
Therefore, the first step is to calculate the current flowing through the inductor:
I = 12 V / 100 Ω
I = 0.12 A
Now that we know the current flowing through the inductor, we can calculate the magnetic field magnitude inside the inductor:
B = (4π × 10⁻⁷ T m/A) * (10,000 turns/m) * (0.12 A) / L
Unfortunately, the length of the solenoid is not provided in the question. To determine the magnetic field magnitude inside the inductor, we need to know the length of the solenoid.