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 negative rate of change of magnetic flux.
First, let's find the rate of change of magnetic flux. The magnetic field inside the solenoid can be calculated using the formula:
B = μ₀ * n * I
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 per unit length (600 turns/m), and I is the current (0.7 A).
Substituting the given values, we can find the magnetic field as:
B = (4π × 10^-7 T*m/A) * (600 turns/m) * (0.7 A)
B = 5.28 × 10^-4 T
Now, let's find the initial and final magnetic flux through the coil. The initial flux is given by:
Φ_initial = B * A
where A is the cross-sectional area of the solenoid (2.60 × 10^-4 m^2).
Substituting the values, we have:
Φ_initial = (5.28 × 10^-4 T) * (2.60 × 10^-4 m^2)
Φ_initial = 1.3728 × 10^-7 Wb
Since the current in the solenoid dies to zero, the final magnetic flux becomes zero as well:
Φ_final = 0
Now, we can calculate the rate of change of magnetic flux by taking the difference between the initial and final flux values and dividing it by the time taken:
Rate of change of magnetic flux (ΔΦ/Δt) = (Φ_final - Φ_initial) / t
= (0 - 1.3728 × 10^-7 Wb) / 0.11 s
= -1.25 × 10^-6 Wb/s
Finally, we can use Faraday's law to find the induced emf in the coil:
emf = -N * (ΔΦ/Δt)
where N is the number of turns in the coil (10 turns). Substituting the values, we get:
emf = - (10 turns) * (-1.25 × 10^-6 Wb/s)
emf = 1.25 × 10^-5 V
Since the resistor is connected to the ends of the coil, the average current induced in the coil can be found by dividing the induced emf by the resistance:
Average current induced (I_induced) = emf / R
= (1.25 × 10^-5 V) / (0.8 Ω)
= 1.5625 × 10^-5 A or 15.625 μA
Therefore, the average current induced in the coil is 1.5625 × 10^-5 A or 15.625 μA.