In a magnetic field of `0.05T`, area of a coil changes from `101 cm^(2)` to `100 m^(2)` without changing the resistance which is `2 Omega`. The amount of charge that flow during this period is

Options

A. `2.5xx10^(-6)` coulomb
B. `2xx10^(-6)` coulomb
C. `10^(-6)` coulomb
D. `8xx10^(-6)` coulomb

Answer

A

Explanation

`varphi=BA`
`implies`change in flux `dvarphi=B.dA=0.05(101-100)10^(-4)0.10^(-6)Wb`.
Now, charge `dQ=(dvarphi)/(R )=(5xx10^(-6))/(2)=2.5xx10^(-6)C`.