A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 50 rad. `s^(-1)` in a uniform horizontal magnetic field of magnitude `3.0xx10^(-2)T`. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance `10 Omega`, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Explanation

Flux through each turn of the coil,
`phi=pir^(2)Bcos(omegat)`
`therefore e=-N(dphi)/(dt)=-Npir^(2)B(d)/(dt)cosomegat=Npir^(2)Bomegasinomegat`
`therefore e_(max)=Npir^(2)Bomega`
`=20xx3.14xx(8xx10^(-2))^(2)xx3xx10^(-2)xx50`
`=0.603V`
The average emf over a complete cycle is zero.
`because I_(max)=(e_(max))/(r)=(0.603)/(10)=0.0603A`
Average dissipated power `=(1)/(2)e_(max)xxI_(max)`
`=(1)/(2)xx0.603xx0.0603=0.018W`
The induced emf produces a torque which opposes the motion of the coil. An external machine is to be used which will produce an equal and opposite torque so as to maintain the motion of the coil. This machine supplies the power apart of which is being transferred to heat.