A coil with resistance 8 `Omega` and having 8 turns is connected with a galvanometer having resistance 8 times the resistance of coil. In 4 ms, if magnetic flux linked with this loop changes from `12xx10^(-5)` Wb to `18xx10^(-5)` Wb then current induced in the loop will be ......

Options

A. 1.6 A
B. `1.6xx10^(-6)` A
C. `1.6xx10^(-3)` A
D. `1.6xx10^(-4)` A

Answer

C

Explanation

Here `Deltaphi=phi_2-phi_1=18xx10^(-5)-12xx10^(-5)`
`=6xx10^(-5)` Wb
N=8
`R=8 Omega, G=8xx8=64 Omega`
`t=4xx10^(-3)` s
`therefore` Induced current,
`I=epsilon/R` ,
`=(N Delta phi)/(Deltat)xx1/(R+G)` [Where R.=R+G]
`=(8xx6xx10^(-5))/(4xx10^(-3)xx72)`
`=0.166xx10^(-2)`
`therefore I approx 1.6xx10^(-3)` A