In the given arrangement the square loop of area `10 cm^(2)` rotates with an angular velocitys `omega` about its diagonal. The loop is connected to a inductance of `L = 100 mH` and a capacitance of `10 mF` in series. The lead wires have a net resistance of `10 Omega`. Given that `B = 0.1 T` and `omega = 63 "rad"//s`

Find the rms current

Options

A. `6.12 xx 10^(-6) J`
B. `8.12 xx 10^(-5) J`
C. `5.12 xx 10^(-5) J`
D. `8.12 xx 10^(6) J`

Answer

B

Explanation

`E = V_("rms") i_("rms") cos phi t = ((B^(2) A^(2) omega^(2))/(2Z)) ((R )/(Z)) (50)`
`= ((0.1)^(2) (10^(-3))^(2) (63)^(2) (10) (50))/(2{(63 xx 0.1 - (1)/(63 xx 0.01))^(2) + (10)^(2)})`
`8.12 xx 10^(-5) J`