Complete Question
A very long, straight solenoid with a cross-sectional area of 2.39cm2 is wound with 85.7 turns of wire per centimeter. Starting at t= 0, the current in the solenoid is increasing according to i(t)=( 0.162A/s2) t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
The value is [tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 2.39 \ cm^2 = \frac{2.39}{10000} = 0.000239 \ m^2[/tex]
The number of turns is [tex]N = 85.7 \ turns/cm = 8570 \ turns / m[/tex]
The initial time is t = 0s
The current on the solenoid is [tex]I(t) = (0.162 \ A/s^2) t^2[/tex]
The number of turns of the secondary winding is [tex]n = 5 \ turns[/tex]
Generally At I = 3.2 A
[tex]3.2 = (0.162 )t^2[/tex]
=> [tex]t^2 = 19.8[/tex]
=> [tex]t = 4.4 \ s[/tex]
Generally induced emf is mathematically represented as
[tex]\epsilon = A * \mu_o * n * N \frac{d(I)}{dt}[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2t[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2 * 4.4[/tex]
[tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
