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

`(dI)/(dt) = 2.59\ x\ 10^4\ A/s`

Step-by-step explanation:

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

`L = (\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 = ((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(dI)/(dt)\\\\ (dI)/(dt) =(emf)/(L) \\\\ (dI)/(dt) =(600\ V)/(0.0232\ H)\\\\ (dI)/(dt) = 2.59\ x\ 10^4\ A/s`