Question

A very long, straight solenoid with a cross-sectional area of 2.22 cm2 is wound with 85.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)=(0.177A/s2)t2. A secondary winding of 5.0 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 1

The induce emf is
`\epsilon = 1.7966*10^(-5) V`

Step-by-step explanation:

From the question we are told that

The cross-sectional area is
`A = 2,22 cm^3 = (2.22)/(10000) = 2.22*10^(-4) \ m^2`

The number of turn is
`N = 85.6 \ turns/cm = 85.6 \ (turns )/((1)/(100) ) = 8560 \ turns / m`

The starting time is
`t_o` = 0 s

The current increase is
`I(t) = (0.177A/s^2) t^2`

The number of turn of secondary winding is
`N_s = 5 \ turn s`

The current at the solenoid is
`I_(t) = 3.2 \ A`

at
`I_(t) = 3.2 \ A`

`3.2 = 0.177* t^2`

=>
`t = \sqrt{ (3.2)/(0.177) }`

`t = 4.25 s`

Generally Faraday's law of induction is mathematically represented as

`\epsilon = A\mu_o N_s N * (di)/(dt)`

`\epsilon = A\mu_o N_s N * (d (0.177 t^2))/(dt)`

`\epsilon = A\mu_o N_s N * (0.177)(2t)`

substituting values

`\epsilon = (2.22*10^(-4)) * ( 4\pi * 10^(-7)) * 5 * [8560]* 0.177 * 2 * 4.25`

`\epsilon = 1.7966*10^(-5) V`