A solenoid has a cross-sectional area of 2.60 10-4 m2, consists of 600 turns per meter, and carries a current of 0.7 A. A 10 turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 0.8 resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.11 s. Find the average current induced in the coil.
To find the average current induced in the coil, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a circuit is equal to the rate of change of magnetic flux through the circuit.
The magnetic flux through the coil can be calculated by multiplying the magnetic field strength (B) and the cross-sectional area (A) of the solenoid. The magnetic field strength inside a solenoid can be approximated as B = μ₀nI, where μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), n is the number of turns per unit length (600 turns/m), and I is the current in the solenoid (0.7 A).
So, the magnetic field strength inside the solenoid is B = (4π × 10⁻⁷ T·m/A) × (600 turns/m) × (0.7 A) = 5.283 × 10⁻⁴ T.
Next, we need to calculate the change in magnetic flux through the coil. Since the number of turns in the coil is 10 and the magnetic field strength is uniform inside the solenoid, the change in magnetic flux is simply the product of the magnetic field strength and the area of the coil: ΔΦ = B × A.
The area of the coil is equal to the circumference of the solenoid multiplied by the number of turns in the coil: A = (2πr) × N, where r is the radius of the solenoid.
Assuming the solenoid is tightly wrapped, the radius of the solenoid is approximately the same as the radius of the coil. Therefore, the area of the coil is A = (2πr) × N = (2π) × (r × N) = (2π) × (r × 10), where N is the number of turns in the coil.
Now we can calculate the change in magnetic flux: ΔΦ = B × A = (5.283 × 10⁻⁴ T) × [(2π) × (r × 10)].
Substituting the given values, we have ΔΦ = (5.283 × 10⁻⁴ T) × [(2π) × (r × 10)].
Finally, using Faraday's law, we can calculate the average induced current in the coil. The induced emf is equal to the rate of change of magnetic flux through the coil, divided by the time taken for the change: ε = ΔΦ/Δt.
Since the resistance connected to the coil is 0.8 Ω, Ohm's law tells us that the induced current is given by I = ε/R, where R is the resistance.
Substituting the given values, we have I = (ΔΦ/Δt) / R = [(5.283 × 10⁻⁴ T) × [(2π) × (r × 10)]] / (0.11 s × 0.8 Ω).
By plugging in the values of the radius (r) and solving the equation, we can find the average current induced in the coil.
To find the average current induced in the coil, we can use Faraday's Law of electromagnetic induction.
Faraday's Law states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through a coil. In equation form, it can be written as:
EMF = -N * Δφ/Δt
where EMF is the induced electromotive force, N is the number of turns of the coil, Δφ is the change in magnetic flux, and Δt is the change in time.
First, let's calculate the change in magnetic flux (Δφ). The magnetic flux through a solenoid is given by:
φ = B * A
where B is the magnetic field strength and A is the cross-sectional area of the solenoid.
Given that the cross-sectional area is 2.60 * 10^-4 m^2, we need to find the magnetic field (B).
The magnetic field inside a solenoid can be calculated using the equation:
B = μ₀ * N₁ * I₁
where μ₀ is the permeability of free space (4π * 10^-7 Tm/A), N₁ is the number of turns per meter of the solenoid, and I₁ is the current in the solenoid.
Given that the solenoid has 600 turns per meter and a current of 0.7 A, we can substitute these values into the equation to find B:
B = (4π * 10^-7 Tm/A) * 600 turns/m * 0.7 A
B ≈ 5.3 * 10^-4 T
Now, we can substitute the values of B and A into the equation φ = B * A to find the initial magnetic flux (φ):
φ = (5.3 * 10^-4 T) * (2.60 * 10^-4 m^2)
φ ≈ 1.4 * 10^-7 Wb
Next, we can calculate the change in magnetic flux (Δφ). Since the current in the solenoid dies to zero, the final magnetic flux is zero:
Δφ = φ - 0
Δφ = 1.4 * 10^-7 Wb - 0
Δφ ≈ 1.4 * 10^-7 Wb
Given that the time for the current in the solenoid to die to zero is 0.11 s, we can substitute the values of Δφ and Δt into Faraday's Law to find the induced EMF:
EMF = -N * Δφ/Δt
EMF = -10 turns * (1.4 * 10^-7 Wb)/(0.11 s)
EMF ≈ -1.3 * 10^-5 V
Since the ends of the coil are connected to a 0.8 Ω resistor, we can use Ohm's Law (V = I * R) to find the average current induced in the coil:
I = EMF/R
I = (-1.3 * 10^-5 V)/(0.8 Ω)
I ≈ -1.6 * 10^-5 A
Therefore, the average current induced in the coil is approximately -1.6 * 10^-5 A. Note that the negative sign indicates that the induced current is in the opposite direction to the original current in the solenoid.