Question

1. A 2,000-turn solenoid is 65 cm long and has cross-sectional area 30 cm2. What rate of change of current will produce a 600 Volts emf in this solenoid.

Answer 1

Answer:

$\frac{dI}{dt} = 2.59\ x\ 10^4\ A/s$

Explanation:

First, we will calculate the inductance of the solenoid by using the following formula:

$L = \frac{\mu_o AN^2}{l}$

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,

$L = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(3\ x\ 10^{-3}\ m^2)(2000)^2}{0.65\ m}\\\\L = 0.0232\ H$

Now, we will use Faraday's law to calculate the rate of change of current:

$emf = L\frac{dI}{dt}\\\\ \frac{dI}{dt} =\frac{emf}{L} \\\\ \frac{dI}{dt} =\frac{600\ V}{0.0232\ H}\\\\ \frac{dI}{dt} = 2.59\ x\ 10^4\ A/s$