Answer: 0.43 V
Explanation:
L = [μ(0) * N² * A] / l
Where
L = Inductance of the solenoid
N = the number of turns in the solenoid
A = cross sectional area of the solenoid
l = length of the solenoid
7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24
1.752*10^-3 = 4π*10^-7 * 202500 * A
1.752*10^-3 = 0.255 * A
A = 1.752*10^-3 / 0.255
A = 0.00687 m²
A = 6.87*10^-3 m²
emf = -N(ΔΦ/Δt).........1
L = N(ΔΦ/ΔI) so that,
N*ΔΦ = ΔI*L
Substituting this in eqn 1, we have
emf = - ΔI*L / Δt
emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3
emf = 0.0234 / 0.055
emf = 0.43 V
The cross-sectional area of the solenoid is 6.88 x 10⁻³ m².
The induced emf in the solenoid when the current drops is 0.425 V.
The given parameters;
The cross-sectional area of the solenoid is calculated as follows;
[tex]L = \frac{\mu_o N^2 A}{l} \\\\A = \frac{Ll}{\mu_0 N^2} \\\\A = \frac{(7.3\times 10^{-3}) \times (0.24)}{4\pi \times 10^{-7} \times 450^2} \\\\A = 6.88\times 10^{-3} \ m^2[/tex]
The induced emf in the solenoid is calculated as follows;
[tex]emf = L \frac{dI}{dt} \\\\emf = (7.3\times 10^{-3})\times \frac{(3.2 -0)}{55\times 10^{-3}} \\\\emf = 0.425 \ V[/tex]
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