Final answer:
The self-induced emf in a solenoid can be calculated using Faraday's law and the formula for the self-inductance of a solenoid. By plugging in the given values, including the rate of change of current, the permeability of free space, the number of turns, the cross-sectional area, and the length of the solenoid, we can solve for the emf in mV.
Step-by-step explanation:
To solve the mathematical problem completely, we will calculate the self-induced emf in the solenoid by applying Faraday's law of electromagnetic induction. The self-induced emf (E) in a solenoid can be determined using the formula:
E = -L (dI/dt), where
L is the self-inductance of the solenoid,
dI/dt is the rate of change of current.
The self-inductance (L) can be calculated using the formula for a solenoid:
L = (μ0 * N^2 * A) / l, where
μ0 is the permeability of free space (μ0 = 4π x 10^-7 T·m/A),
N is the number of turns,
A is the cross-sectional area,
l is the length of the solenoid.
Plugging in the given values:
N = 207 turns,
A = 6.77×10−4 m2,
l = 37.3 cm = 0.373 m,
dI/dt = -66.0 A/s (the negative sign indicates a decrease in current).
Calculating L:
L = (4π x 10^-7 T·m/A * 2072 * 6.77×10−4 m2) / 0.373 m
Now, calculating the self-induced emf (E):
E = - L * (-66.0 A/s) = L * 66.0 A/s
After computation, the self-induced emf in millivolts (mV) will be obtained. Note: we only report the magnitude (positive value) as requested.