suppose the ends of the coil in the previous problem are connected to a resistane of `100 Omega`. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.

Explanation

Since`H=int_(0)^(1 min) i^2 R.dt`
`=int_(0)^(1min) (B^2A^2omega^2)/(R^2) sin omega t.R dt`
`=(B^2 A^2 omega^2)/(2R). Int_(0)^(1 min) (1-cos 2 omega t).dt`
`=(B^2A^2omega^2)/(2R)(1-(sin 2 omega t)/(2 omega))_(0)^(1 min)`
`(B^2 A^2 omega^2)/(2R)(60-(sin.2xx80xx2 pi//60xx60)/(2xx80xx2 pi//60))`
`=(60)/(2R)xxpi^2r^4xxB^2xx(80xx(2 pi)/(60))^2`
``
`= 60/200xx10xx64/9xx10xx625xx10^(-8)xx10^(-4)`
`=(625xx6xx64)/(9xx2)xx10^(-11)=1.33xx10^(-7)J.`