A current l = 1.5 A is flowing through a long solenoid of diameter 3.2 cm, having 220 turns per cm. At its centre, a 130 turn closely packed coil of diameter 2.1 cm is placed such that the coil is coaxial with the long solenoid. The current in the solenoid is reduced to zero at a steady rate in 25 ms. What is the magnitude of emf induced in the coil while the current in the solenoid is changing?

Explanation

Because coil C consists of more than one turn, we apply Faraday.s law in the form of Eq. 30-5 `(E = -N dphi_(g)//dt)`, where the number ofturns N is 130 and `dphi_(B)//dt` is the rate at which the flux changes.
Because the current in the solenoid decreases at a steady rate, flux `phi_(B)` also decreases at a steady rate, and so we can write `dphi_(B)//dt` as `Deltaphi_(B)//Deltat`. Then, to evaluate `Deltaphi_(B)` , we need the final and initial flux values. The final flux `Deltaphi_(B.f)` is zero because the final current in the solenoid is zero. To find the initial flux `wedgephi_(B.P)`, we note that area A is `1//4pid^(2)= (3.464 xx 10^(-4)m^(2))` and the number n is 220 turns/cm, or 22 000 turns/m. Substituting Eq. 29-20 into Eq. 30-2 then leads to
`phi_(B.i)=BA=(mu_(0)"in")A`
`=(4pixx10^(-7)T.m//A)(1.5A)(2200 "turns/m")`
`xx(3.464xx10^(-4)m^(2))`
`=1.44xx10^(-5)Wb`.
Now we can write
`(dphi_(B))/(dt)=(Deltaphi_(B))/(Deltat)=(phi_(B.f)-phi_(B.i))/(Deltat)`
`=((0-1.44xx10^(-5)Wb))/(25xx10^(-3)s)`
`=-5.76xx10^(-4)Wb//s=-5.76xx10^(-4)V`
We are interested only in magnitudes, so we ignore the minus signs here and in Eq. 30-5, writing
`E=N(dphi_(B))/(dt)`=(130 turns)`(5.76xx10^(-4)V)`
`=7.5xx10^(-2)V`
=75 mV