The linear loop has an area of `5 xx 10^(-4)m^(2)` and a resistance of `2 Omega`. The larger circular loop is fixed and has a radius of `0.1m`. Both the loops are concentric and co-planner. The smaller loop is rotated with an angular velocity `omegarads^(-1)` about its dismeter. The magnetic flux with the smaller loop is

Options

A. `2pi xx 10^(-6)Wb`
B. `pi xxx 10^(-9)Wb`
C. `pi xx 10^(-9)cos omegat Wb`
D. zero

Answer

C

Explanation

( c) Magnetic field due to larger loop
`= (mu_(0)1)/(2r) = (4pi xx 10^(-7) xx 1)/(2 xx 0.1) T = 2pi xx ^(10^(-6) T` ltbr. Now, magnetic flux linked with smaller loop, `phi`
`= NBA cos omegat = 1 xx 2pi xx 10^(-6) xx 5 xx 10^(-4) cos omegat`
`= pi xx 10^(-9) cos omegat weber`