Answer:
[tex]\frac{dI}{dt} = 2.59\ x\ 10^4\ A/s[/tex]
Explanation:
First, we will calculate the inductance of the solenoid by using the following formula:
[tex]L = \frac{\mu_o AN^2}{l}[/tex]
where,
L = self-inductance of solenoid = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
A = Cross-sectional area = 30 cm² = 3 x 10⁻³ m²
N = No. of turns = 2000
l = length = 65 cm = 0.65 m
Therefore,
[tex]L = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(3\ x\ 10^{-3}\ m^2)(2000)^2}{0.65\ m}\\\\L = 0.0232\ H[/tex]
Now, we will use Faraday's law to calculate the rate of change of current:
[tex]emf = L\frac{dI}{dt}\\\\ \frac{dI}{dt} =\frac{emf}{L} \\\\ \frac{dI}{dt} =\frac{600\ V}{0.0232\ H}\\\\ \frac{dI}{dt} = 2.59\ x\ 10^4\ A/s[/tex]
