A tong solenoid with 15 turns per em has a small loop of area `2.0 cm^(2)` placed inside the solenoid normal to its axis, If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the inducced emf in the loop while the current is changing?

Explanation

Given number of turns (n) =15 per cm = 1500 per metre.
Area of small loop `A=2cm^(2)=2xx10^(-4)m^(2)`
Change in current `(dI)/(dt)=(4-2)/(0.1)=(2)/(0.1)=20A//s`
Let e be the induced emf, According to Faraday's law
`e=(dphi)/(dt)=(d)/(dt)(BA)or e=A.(dB)/(dt)" "(therefore phi=BA)`
`e=A.(d)/(dt)(mu_(0)nI) (therefore "Magnetic field inside the solenoid B"=mu_(0)NI)`
`or e=Amu_(0) (dI)/(dt)`
`e=2xx10^(-4)xx4xx3.14 xx 10^(-7)xx1500xx20`
`e=7.5xx10^(6)v(therefore mu_(0)=4pixx10^(-7))` Thus, the inducced emf in the loop is `7.5 xx 10^(6)V`.