Question

A long solenoid with 15 turns per cm has small loop of area `2.0 cm^(2)` placed inside, normal to the axis of the soleniod. If current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 s, what is the induced voltage in the loop, while the current is changing ?

Answer 1

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 `cm^(2)` = `2 × 10^(−4) m^(2)`
Current carried by the solenoid changes from 2 A to 4 A.
`therefore`Change in current in the solenoid, di = 4 − 2 = 2 A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday’s law as:
`e=(dphi)/(dt)`....(i)
Where,
`phi`= Induced flux through the small loop
= BA ... (ii)
B = Magnetic field
=`mu_(0)ni`.....(iii)
`mu_(0)` = Permeability of free space
= `4pi×10^(−7)` H/m
Hence, equation (i) reduces to:
`e=d/(dt)(BA)`
`=Amu_(0)nxx((di)/(dt))`
`=2xx10^(-4)xx4pixx10^(-7)xx1500xx2/0.1`
`=7.54xx10^(-6)` V
Hence, the induced voltage in the loop is `=7.54xx10^(-6)`