Question

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of `50" rad s"^(-1)` in a uniform horizontal magnetic field of magnitude `3.0 xx10^(-2)T` Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 100, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from ?

Answer 1

Given radius of coil =8cm =0.08m, Number of turns=20
Resistance of closed loop `=10Omega,` Angular speed `omega= 50"rad/s"`
Magnitude of magnetic field `B =3 xx 10^(-2)T`
Induced emf produced in the coil `e="NBA W "sin omegat` For maximum emf, `sin omegat= 1`
`therefore` Maximum emf `e_0=NBAomega`
`e_0=20xx3xx10^(-2)xx3.14xx(0.08)^2xx50 Rightarrow e_(0)=0.603V`
Maximum current in the coil `I_(0)=(e_(0))/(R)=(0.603)/(10)=0.603a`
Average induced emf `e_(av)=1/T underset(0)overset(2pi)intedt=1/Tunderset(0)overset(2pi)int NBA omegasin omegat dt`
`e_(av)=1/T.NAB omega[(cos omegat)/(omega)]_(0)^(2pi) Rightarrow e_(Av)=(NBA)/(T)[cos 2pi-cos 0^(@)] Rightarrow e_(av)=(NBA)/(T)(1-1)=0`
For full cycle average emf `e_(av)=0`
Average power loss due to heating `=(E_(0)I_(0))/(2)=(0.603xx0.603)/(2)=0.018W`
The source of power dissipated as heat in the coil is the external rotar. The current induced in the coil causes a torque which opposses the rotation of the coil, so the external agent rotar counter this torque to keep the coil rotating uniformly.